RBSE Solutions for Class 10 Maths Chapter 18 Probability Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 18 Probability Additional Questions.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 18 |
| Chapter Name | Probability |
| Exercise | Additional Questions |
| Number of Questions Solved | 59 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 10 Maths Chapter 18 Probability Additional Questions
Multiple Choice Questions
Question 1.
An event cannot be occured, then it’s (RBSESolutions.com) probability is : [NCERT Exemplar Problem]
(A) 1
(B) \(\frac { 3 }{ 4 }\)
(C) \(\frac { 1 }{ 2 }\)
(D) 0
Solution :
The option (D) is correct.
Question 2.
There are very few chances of occuring an event. It’s probability is nearly :
(A) 0.0001
(B) 0.001
(C) 0.01
(D) 0.1
Solution :
The option (A) is correct.
Question 3.
On expressing the probability of occuring an event in the (RBSESolutions.com) terms of percentage. It can never be taken place. [NCERT Exemplar Problem]
(A) less than 100
(B) less than 0
(C) more than 1
(D) Besides a whole number everything
Solution :
The option (B) is correct.
Question 4.
If P(A)shows the probability of occuring an event, then : [NCERT Exemplar Problem]
(A) P(A) < 0
(B) P(A) > 1
(C) 0 ≤ P(A) ≤ 1
(D) -1 ≤ P(A) ≤ 1
Solution :
The option (C) is correct.
Question 5.
A card is drawn from a deck of 52 playing cards at random. The (RBSESolutions.com) probability that the drawn card is a face card in red colour is:
(A) \(\frac { 3 }{ 26 }\)
(B) \(\frac { 3 }{ 13 }\)
(C) \(\frac { 2 }{ 13 }\)
(D) \(\frac { 1 }{ 2 }\)
Solution :
Number of playing cards in a deck = 52
∴ Number of all possible outcomes = 52
The red face card in a deck = 6
∴ The number of favourable outcomes = 6
The probability of drawing a red face card from
the deck = \(\frac { 6 }{ 52 }\) = \(\frac { 3 }{ 26 }\)
Hence, the option (A) is correct.
Question 6.
A card is drawn from a well shuffled deck of 52 playing cards. The probability that the card drawn not an ace of diamond is E. Then number of favourable outcomes of E is : [NCERT Exemplar Problem]
(A) 4
(B) 13
(C) 48
(D) 51
Solution :
The number of playing cards in a deck = 52.
One card is drawn from this well shuffled deck.
The probability that the card drawn is an ace of diamond is E.
∴ The number of favourable outcomes of E = 52 – 1 = 51
Hence, the option (D) is correct.
Question 7.
A girl estimated that the probability of winning first (RBSESolutions.com) prize of lottery is 0.08. If over all 6000 tickets of the lottery are sold then tickets the girl buy : [NCERT Exemplar Problem]
(A) 40
(B) 240
(C) 480
(D) 750
Solution :
Number of tickets sold = 6000
∴ The number of all possible outcomes = 6000
Let the number of tickets bought by the girl be x.
∴ The number of favourable outcomes = x
Probability of winning first prize
⇒ 0.08 = \(\frac { x }{ 6000 }\)
⇒ x = 0.08 × 6000
⇒ x = 480
Hence, the option (C) is correct.
Question 8.
The probability of “not a black jack” in well (RBSESolutions.com) shuffled deck of 52 playing cards will be:
(A) \(\frac { 1 }{ 26 }\)
(B) \(\frac { 3 }{ 4 }\)
(C) \(\frac { 11 }{ 52 }\)
(D) \(\frac { 25 }{ 26 }\)
Solution :
The number of playing cards in a deck = 52
∴ Number of all possible outcomes 52
Black jack in a deck are spades and clubs
∴ Number of favourable outcome =2
The probability that a black jack in the deck
= \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
∴ The probability that not blackjack in the deck
= 1 – \(\frac { 1 }{ 26 }\) = \(\frac { 25 }{ 26 }\).
Hence. the option (D) is correct.
Question 9.
A bag contains cards with numbered 2, 3, 4, …… 11 on each of them (RBSESolutions.com) respectively. A card is drawn at random. The probability that the drawn card has a prime number on it is. (CBSE 2012)
(A) \(\frac { 1 }{ 2 }\)
(B) \(\frac { 2 }{ 5 }\)
(C) \(\frac { 3 }{ 10 }\)
(D) \(\frac { 5 }{ 9 }\)
Solution :
The printed numbers on the cards are 2, 3, 4, 11.
∴ The number of cards in the bag = 10
The prime numbers printed on the cards are = 2, 3, 5, 7, 11
The number of favourable outcomes = 5
∴ The probability that the card drawn has a prime number = \(\frac { 5 }{ 10 }\) = \(\frac { 1 }{ 2 }\)
Hence, the option (A) is correct.
Question 10.
A bag contains some cards. The numbers 6 – 50 are printed on them. A card is at (RBSESolutions.com) random drawn from the bag. The probability that the number drawn is a perfect square is:
(A) \(\frac { 1 }{ 15 }\)
(B) \(\frac { 2 }{ 15 }\)
(C) \(\frac { 1 }{ 9 }\)
(D) \(\frac { 4 }{ 45 }\)
Solution :
The numbers 6 to 50 are printed on the cards in the bag.
∴ The number of all possible outcomes = 45
The perfect square numbers printed on the cards are 9, 16, 25, 36, 49
The number of favourable outcomes = 5
The required probability = \(\frac { 5 }{ 45 }\) = \(\frac { 1 }{ 9 }\)
Hence, the option (C) is correct.
Question 11.
A dice is thrown once. The probability that the top (RBSESolutions.com) face of dice contains a prime number is : (CBSE 2013)
(A) \(\frac { 2 }{ 3 }\)
(B) \(\frac { 1 }{ 3 }\)
(C) \(\frac { 1 }{ 2 }\)
(D) \(\frac { 1 }{ 6 }\)
Solution :
A dice is thrown once, the numbers that may be on the top face are 1, 2, 3, 4, 5, 6.
∴ The number of all possible out comes = 6
And the primes numbers are 2, 3 and 5.
∴ The number of favourable outcomes = 3.
∴ The required probability = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
Hence, the option (C) is correct.
Question 12.
There are five houses A, B, C, D and E in a school. There are 23 students (RBSESolutions.com) in a class, in which 4 students in house A, Sin house B, 5 in house C, 2 in house D and remaining students are in the house E. Out of these one students is selected for being monitor. The probability that the student chosen is not of the houses A, B and C is : (NCERT Exemplar Problem)
(A) \(\frac { 4 }{ 23 }\)
(B) \(\frac { 6 }{ 23 }\)
(C) \(\frac { 8 }{ 23 }\)
(D) \(\frac { 17 }{ 23 }\)
Solution :
The total number of students in the class = 23
The sum of the students of the houses A, B and C = 4 + 8 + 5 = 17
The number of all possible outcomes = 23
The number of favourable outcomes = 17
The probability that the student choosen from the houses A, B and C = \(\frac { 17 }{ 23 }\)
The probability that the ‘student not’ from the houses if A, B and C.
= 1 – \(\frac { 17 }{ 23 }\) = \(\frac { 6 }{ 23 }\)
Hence, the option (B) is correct.
Question 13.
If two coins are tossed simultaneously, the (RBSESolutions.com) probability of getting two heads is :
(A) \(\frac { 3 }{ 4 }\)
(B) \(\frac { 1 }{ 2 }\)
(C) \(\frac { 1 }{ 4 }\)
(D) 1
Solution :
When two coins are tossed simultanously the all possible outcomes are HH, HT, TH, TT
The number of all possible outcomes = 4
The number of favourable outcomes of getting two heads = 1
∴ The probability of getting two heads = \(\frac { 1 }{ 4 }\)
Hence, the option (C) is correct.
Question 14.
When a dice is thrown, the probability of getting an odd (RBSESolutions.com) number smaller than 3. (NCERT Exemplar Problem)
(A) \(\frac { 1 }{ 6 }\)
(B) \(\frac { 1 }{ 3 }\)
(C) \(\frac { 1 }{ 2 }\)
(D) 0
Solution :
When a dice is thrown once then the number of all possible outcomes = 6
The odd number smaller than 3 = 1
The probability of getting an odd number smaller than 3 = \(\frac { 1 }{ 6 }\)
Hence, the option (A) is correct.
Question 15.
The numbers 1 to 12 are written on the 12 tickets respectively. One (RBSESolutions.com) ticket is drawn at random. The probability that the number on the drawn ticket is a multiple of 3 is :
(A) \(\frac { 2 }{ 3 }\)
(B) \(\frac { 1 }{ 12 }\)
(C) \(\frac { 1 }{ 2 }\)
(D) \(\frac { 1 }{ 3 }\)
Solution :
Multiples of 3 are 3, 6, 9 and 12 between the numbers 1 to 12.
Total number of all possible outcomes = 12
The number of multiples of 3 = 4
∴ The probability that the number drawn is a multiples of 3 = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
Hence, the option (D) is correct.
Question 16.
Two dice are thrown simultaneously. The probability that the (RBSESolutions.com) two top faces of dice has same number is:
(A) \(\frac { 1 }{ 6 }\)
(B) \(\frac { 5 }{ 6 }\)
(C) \(\frac { 1 }{ 36 }\)
(D) None of these
Solution :
When two dice are thrown at the same time, then the total number of all possible outcomes = 6 × 6 = 36
The favourable outcomes of getting same digits on the top faces of two dice are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
∴ The number of favourable outcomes = 6.
The required probability = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
Hence, the option (A) is correct.
Question 17.
Three coins are tossed simultaneously. The probability of (RBSESolutions.com) getting only one head will be :
(A) \(\frac { 1 }{ 2 }\)
(B) \(\frac { 3 }{ 8 }\)
(C) \(\frac { 5 }{ 8 }\)
(D) \(\frac { 1 }{ 8 }\)
Solution :
The related events of tossing three coins simultaneously are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
So the number of all possible outcomes = 8
The number of getting one head = 3.
∴ The required probability = \(\frac { 3 }{ 8 }\)
Hence, the option (B) is correct.
Question 18.
There are 3 red and 4 blue marbles in a bag. At random, one marble is drawn. The (RBSESolutions.com) probability that the marble drawn is blue :
(A) 1
(B) \(\frac { 3 }{ 7 }\)
(C) \(\frac { 4 }{ 7 }\)
(D) \(\frac { 2 }{ 7 }\)
Solution :
Total number of marbles in the bag = 3 + 4 = 7
∴ The number of all possible outcomes = 7
The number of favourable outcomes of getting blue marbles = 4
∴ Required probability = \(\frac { 4 }{ 7 }\)
Hence, the option (C) is correct.
Very Short Answer Type Questions
Question 1.
Write the sum of probabilities of all primary (RBSESolutions.com) events of an experiment.
Solution :
The sum of probabilities of all primary events of an experiment is 1.
Question 2.
A bag contains 3 red and 5 black balls. One ball is taken out from this bag at random. What is the probability that ball drawn is black?
Solution :
Number of balls in the bag = 3 red + 5 black.
∴ Total number of all possible outcomes = 8.
The number of favourable outcomes that the ball is not black = 3.
∴ The required probability = \(\frac { 3 }{ 8 }\)
Question 3.
If a bag contains 4 balls of different colours Red (R), Blue (B), Yellow (Y) and White (W). If two balls (RBSESolutions.com) drawn at random, what is its sample?
Solution :
Two balls are taken out at random which and of different colours. So, the sample set
S = {RB, RY, RW, BY, BW, YW}
Question 4.
There are 12 students in a class in which 5 boys and remaining are girls. If out of them one student is to be choosen, then what is the probability that chosen student is a girl ?
Solution :
Total number of students in the class = 12
The number of boys = 5
And the number of girls = 12 – 5 = 7
When only one student is to be choosen than the the probability that the chosen student is girl = \(\frac { 7 }{ 12 }\)
Question 5.
A dice is thrown once, Find the probability of getting a (RBSESolutions.com) number greater than 3.
Solution :
All possible outcomes of throwing the dice = 1, 2, 3, 4, 5, 6.
∴ The number of all possible outcomes = 6.
The favourable outcomes of getting a number grater than 3 = 4, 5, 6.
∴ Number of favourable outcomes = 3.
∴ Required probability = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
Question 6.
A card is drawn at random from a well shuffled deck of 52 playing cards. Find the probability that the card drawn is spades but not an ace.
Solution :
The deck of 52 playing cards has 13 cards of spades. But it is not an ace, then number of spades = 13 – 1 = 12
∴ Required probability = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)
Question 7.
A card ¡s drawn from a deck of 52 playing cards. Find the probability that the (RBSESolutions.com) card drawn a king or a card of diamonds.
Solution :
A deck of 52 playing cards contains 13 diamond and 3 other kings.
∴ Total number of diamonds cards and 3 kings = 13 + 3 = 16.
The probability that the drawing card is a diamond or a king = \(\frac { 16 }{ 52 }\) = \(\frac { 4 }{ 13 }\)
Question 8.
The numbers 1 to 12 are written on 12 tickets respectively. If out of them one ticket is picked up then find the probability that the ticket picked is a multiple of 2 or 3.
Solution :
Among number 1 to 12, the multiple of 2 or 3 are 2, 3, 4, 6, 8, 9, 10, 12.
∴ The number of all possible outcomes = 12
And the number of favourable outcomes = 8.
∴ Required probability = \(\frac { 8 }{ 12 }\) = \(\frac { 2 }{ 3 }\)
Question 9.
Find the probability of getting a number (RBSESolutions.com) greater than 4, when a dice is thrown.
Solution :
When a dice is thrown, the all possible outcomes = 6.
The number greater than 4, 5 and 6.
The number favourable outcomes = 2.
∴ The required probability = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
Question 10.
In a non-leap year what is the probability of falling 53 Sundays?
Solution :
A non-leap year has 365 days in it in which 52 weeks (RBSESolutions.com) and one extra day. That 1 day may be
Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.
∴ The number of all possible outcomes = 7.
∴ The number of favourable outcomes = 1
∴ The required probability = \(\frac { 1 }{ 7 }\)
Short Answer Type Questions
Question 1.
What is probability that there are 5 Sundays in a month of 30 days, which is chosen at random.
Solution :
A month of 30 days has four complete weeks and 2 extra days. These two days may be;
(i) Monday and Tuesday
(ii) Tuesday and Wednesday
(iii) Wednesday and Thursday
(iv) Thursday and Friday
(v) Friday and Saturday
(vi) Saturday and Sunday
(vii) Sunday and Monday.
∴ The number of all possible events = 7
Now in these events, last two events have Sunday each.
∴ Number of favourable outcomes = 2
∴ In these remaining two days, the probability of having Sunday = \(\frac { 2 }{ 7 }\).
Question 2.
Three coins are tossed simultaneous. Find (RBSESolutions.com) the probability of getting only two heads. (CBSE 2013)
Solution :
Three coins are tossed simultaneously the possible outcomes are.
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Total number of all possible outcomes = 8.
The favourable outcomes of getting 2 heads exactly = HHT, HTH, THH
∴ The number of favourable outcomes = 3
∴ The required probability = \(\frac { 3 }{ 8 }\).
Question 3.
It is given, the probability that two students out of a group of 3 students have not same birthdays is 0.992. What is the probability that they have same birthdays ?
Solution :
The event occurring and not occurring are complementary.
∴ The probability that two students have not same birthdays P(E) = 0.992
∴ The probability that two students have same birthdays P(\(\overline { E }\)) = 1 – 0.992 = 0.008
Question 4.
A card is drawn of random from a well shuffled (RBSESolutions.com) deck of 52 playing cards. Find the probability that the card drawn is:
(i) a diamond
(ii) not a diamond
(iii) a black coloured
(iv) Not a black coloured.
Solution :
The number of playing cards in a deck = 52.
(i) The number of diamonds cards in the deck = 13.
∴ The probability that the card drawn is a diamonds = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(ii) The number of non-diamonds cards = 52 – 13 = 39
∴ The probability that the card drawn is not diamonds = \(\frac { 39 }{ 52 }\) = \(\frac { 3 }{ 4 }\)
(iii) The number of black cards in the deck = 13 spades + 13 clubs = 26.
∴ The probability that the drawn cards is black = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)
(iv) The probability of getting not black card = 1 – \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
Question 5.
Two dice are thrown at a time. Find the probability (RBSESolutions.com) that the top faces of dice have a product 6. (CBSE 2015)
Solution :
When two dice are tossed simultaneously, the total number of all possible outcomes = 6 × 6 = 36.
The favourable outcomes of getting a product 6, on the top surfaces of two dice = (1,6), (2,3), (3, 2), (6, 1)
∴ The number of favourable outcomes = 4
Hence, the required probability = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)
Question 6.
A box contains cards on which the numbers, 3, 5, 7, 9 …….., 35, 37 are (RBSESolutions.com) printed. A card is drawn from the box at random. Find the probability that the card has a prime number. (CBSE 2013)
Solution :
The numbers printed on the cards are 3, 5, 7, 9, 35, 37.
∴ Total number of all possible outcomes = 18
The prime numbers marked on the cards are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37.
The number of favourable outcomes = 11
∴ The required probability = \(\frac { 11 }{ 18 }\)
Question 7.
The probability that a card drawn from a well shuffled deck of 52 playing cards is not an ace.
Solution :
The number of playing cards in deck = 52.
∴ Number of all possible outcomes = 52.
The number of aces in deck = 4.
∴ The number of favourable outcomes = 4
∴ The required probability = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
The probability that the card drawn is not an ace = 1 – \(\frac { 1 }{ 13 }\) = \(\frac { 12 }{ 13 }\)
Question 8.
There are some toffees in a bag with an aroma of lemon. Malini takes (RBSESolutions.com) out a toffee without looking into. What is the probability that the toffee taken out is:
(i) with an aroma of orange
(ii) with an aroma of lemon.
Solution :
Since, the bag contains the toffees only with the aroma of lemon, so if a toffee is taken out at random.
(i) Probability that it has an aroma of orange is zero as all the toffees have lemon flavor.
(ii) Since all the toffees have lemon aroma so the event of taken out of a toffee is said to certain event.
∴ Its probability will be 1.
Hence (i) the probability that the toffee drawn has a orange aroma = 0.
(ii) The probability that the toffee drawn has a lemon aroma = 1.
Question 9.
A card is drawn of random from a well shuffled deck of 52 playing (RBSESolutions.com) cards. Find the probability that the card drawn is a
(i) red king
(ii) A queen or a jack (CBSE 2012)
Solution :
The deck contains 52 playing cards and a card can be drawn by 52 different ways.
∴ Total number of all possible outcomes = 52.
(i) Let the event of getting red king be R.
∴ The number of favourable outcomes of R = 2
P(R) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
(ii) Let the event of getting a queen or jack be A.
∴ The number of favourable outcomes of A = 8.
∴ P(A) = \(\frac { 8 }{ 52 }\) = \(\frac { 2 }{ 13 }\)
Hence (i) P(R) = \(\frac { 1 }{ 26 }\) and (ii) P(A) = \(\frac { 2 }{ 13 }\).
Question 10.
A coin is tossed three times simultaneously. Make a (RBSESolutions.com) table of all possible outcomes.
Also find the probabilities of getting.
(i) all head
(ii) A least two heads. (NCERT Exemplar Problem)
Solution :
A coin is tossed three time so all the possible outcomes are : HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
∴ Total number of all possible outcomes = 8.
(i) Let the event of all heads be A.
The favourable outcomes of A = HHH.
∴ The number of favourable outcomes of (A) = 1.
⇒ P(A) = \(\frac { 1 }{ 8 }\).
(ii) Let the event of getting at least two heads be (B). So the (RBSESolutions.com) favourable outcomes of (B) are HHH, HHT, HTH, THH.
The number of favourable outcomes of B = 4.
⇒ P(A) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
Hence (i) = P(A) = \(\frac { 1 }{ 8 }\)
(ii) P(B) = \(\frac { 1 }{ 2 }\)
Long Answer Type Questions
Question 1.
A jar contains the red, blue and orange balls in it. The probability of drawn a red ball at random is \(\frac { 1 }{ 4 }\). Similarly the probability of drawn a blue ball is \(\frac { 1 }{ 2 }\). If there (RBSESolutions.com) are 10 orange balls in the jar, Find the total number of the balls In the jar in all. (CBSE 2015)
Solution :
Let the probability of taking out the red balls, blue balls and orange balls be E1, E2 and E3 respectively.
We know that P(E1) = \(\frac { 1 }{ 4 }\)
and P(E2) = \(\frac { 1 }{ 3 }\)
∴ P(E3) = 1 – P(E1) – P(E2)
= 1 – \(\frac { 1 }{ 3 }\) – \(\frac { 1 }{ 4 }\) = \(\frac { 5 }{ 12 }\)
We also know that P(E3)
Total number of balls = \(\frac { 120 }{ 5 }\) = 24
Hence, the total number of balls in the jar = 24.
Question 2.
There are 5 red marbles, 8 white marbles and 4 green marbles In a box. A marble is (RBSESolutions.com) drawn at random from this box. Find the probability that the marble drawn is :
(i) red
(ii) white
(iii) not green.
Solution :
Number of red marbles = 5
Number of white marbles = 8
and number of green marbles = 4
The total number of marbles in the box = 5 + 8 + 4 = 17.
When one marble is drawn from the box, then the total number of all possible outcomes = 17
(i) Let the probability of the red marble drawn be R.
∴ The number of favourable outcomes of event R = 5.
∴ P(R)
= \(\frac { 5 }{ 17 }\)
(ii) The number of favourable outcomes of the (RBSESolutions.com) white marble drawn (W) = 8.
∴ The probability that the marble drawn is white P(W)
= \(\frac { 8 }{ 17 }\)
(iii) The number of favourable outcomes that the drawn marble is not green = 5 + 8 = 13.
∴ The probability that the marble drawn ¡s not green.
Number of favourable outcomes
= \(\frac { 13 }{ 17 }\)
Hence (i) P(R) = \(\frac { 5 }{ 17 }\)
(ii) P(W) = \(\frac { 8 }{ 17 }\)
(iii) P(G) = \(\frac { 13 }{ 17 }\)
Question 3.
A card is drawn at random from a well shuffled deck of 52 playing (RBSESolutions.com) cards. Find the probabilities of the following :
(i) A red king
(ii)A face card
(iii)A red face card
(iv) Jack of hearts
(v) A spades card
(vi) The diamonds queen.
Solution :
There are 52 cards in a deck of playing cards. If a card is drawn from this well shuffled deck, the total number of all possible outcomes = 52.
(i) Let the event of drawing a red king be R. Since a deck has four kings in it out of which 2 kings are red.
∴ The number of favourable outcomes of getting a red king.
Let the probability of this event be R =2.
∴ P(R)
= \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
(ii) Let the event according (RBSESolutions.com) a face card be (E)
∵ Since each of the group of has 3 face cards (king, queen, and jack) in it.
∴ Number of face cards in the deck = 3 × 4 = 12
∴ The probability of event (E) = P(E)
= \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)
(iii) Let the event of a red face card be (A)
∵ N number of face cards in the deck = 12
∴ Number of red face cards in the deck = 6
Then the number of possible outcomes of event (A).
⇒ P(A)
= \(\frac { 6 }{ 52 }\) = \(\frac { 3 }{ 26 }\)
(iv) Let the event of hearts jack be (B).
∵ There is only one hearts jack in the deck. So (RBSESolutions.com) the number of favourable outcomes of (B) = 1
∴ The probability of event (B)
⇒ P(B)
= \(\frac { 1 }{ 52 }\)
(v) Let the event occurring a spades card be (C).
∵ The number of spades cards in the deck 13.
∴ The number of favourable outcomes of event (C) = 13.
The probability of event (C) = \(\frac { 13 }{ 52 }\).
⇒ P(C) = \(\frac { 1 }{ 4 }\)
(vi) Let the event occurring a diamonds (RBSESolutions.com) queen be (D).
∵ There is only one diamonds queen in the deck.
∴ The number of favourable outcomes of event (D) = 1
The probability of (D)
⇒ P(D) = \(\frac { 1 }{ 52 }\)
(i) P(R) = \(\frac { 1 }{ 26 }\)
(ii) P(E) = \(\frac { 3 }{ 13 }\)
(iii) P(R) = \(\frac { 3 }{ 26 }\)
(iv) P(R) = \(\frac { 1 }{ 52 }\)
(v) P(R) = \(\frac { 1 }{ 4 }\)
(vi) P(R) = \(\frac { 1 }{ 52 }\)
Question 4.
There are 90 disc in a box, on which numbers 1 to 90 are marked (RBSESolutions.com) respectively. If a disc is draw at random from the box, find the probability that the number is printed on
(i) A number of digits.
(ii) A perfect square number.
(iii) A number divisible by 5.
Solution :
Total number of discs = 90
∴ Total possible outcomes = (1, 2, 3 …….., 90)
The total number of all favourable outcomes = 90.
If 1 dice is drawn at random
(i) Disc with a number of two digits.
∵ The two digits number = (10, 11, 12,13,…..,90)
∴ The number of favourable outcomes = 81
∴ The probability that the disc drawn contains a two digits number
= \(\frac { 81 }{ 90 }\) = \(\frac { 9 }{ 10 }\)
(ii) The perfect squares numbers (RBSESolutions.com) among = (1, 4, 9, 16, 25, 36, 49, 64, 81)
Number of favourable outcomes = 9.
∴ The probability that the disc drawn contains a perfect squares number = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)
The numbers divisible by 5 are 5, 10, 15 …., 90
The number of the numbers divisible by 5 = 18
∴ The number of favourable outcomes = 18
∴ The probability that the disc contains a number divisible by 5 = \(\frac { 18 }{ 90 }\) = \(\frac { 1 }{ 5 }\)
Hence, (i) the probability that the disc contains two digit number = \(\frac { 9 }{ 10 }\)
(ii) The probability that the disc contains a number divisible by 5 = \(\frac { 1 }{ 5 }\)
Question 5.
During a game, a one rupee coin is tossed thrice and obtained outcome is noted. If Haneef gets the same outcomes on the three coins i.e., either heads or tails, he will win the game, otherwise will be defeated. Find the probability of losing the game by Haneef.
Solution :
When a coin of one rupee is tossed three times, the all (RBSESolutions.com) possible outcomes are given as :
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
Total number of all possible outcomes = 8.
Let the probability of getting same outcomes be (A).
The favourable outcomes of occurring same outcomes are HHH, TTT.
∴ The number of favourable outcomes = 2
∴ The probability that Haneef will win the game
P(A) = \(\frac { 2 }{ 8 }\) = \(\frac { 1 }{ 4 }\)
Let the probability of not winning the game be P(\(\overline { A }\))
∴ P(\(\overline { A }\)) = 1 – P(A)
P(\(\overline { A }\)) = 1 – \(\frac { 1 }{ 4 }\) = \(\frac { 3 }{ 4 }\)
Hence, the probability that Haneef will loss the game = \(\frac { 3 }{ 4 }\)
Question 6.
Two customers Shyam and Ekta visit a shop Tuesday to (RBSESolutions.com) Saturday during every week. There visit to the shop or not to visit are at random. Find the probability that both of them will visit the shop.
(i)On the same day
(ii) On consecutive days
(iii) On different days.
Solution :
Two customers Shyam and Ekta visit a shop on Tuesday to Saturday during a week. The visit not visit made by the each customer is at random. All possible outcomes are.
Visit Made by Shyam Possible Visits Made by Ekta
Tuesday Tuesday, Wednesday, Thursday, Friday, Saturday.
Wednesday Tuesday, Wednesday, Thursday, Friday, Saturday.
Thursday Tuesday, Wednesday, Thursday, Friday, Saturday.
Friday Tuesday, Wednesday, Thursday, Friday, Saturday.
Saturday Tuesday, Wednesday, Thursday, Friday, Saturday.
If Tuesday be (T) be Wednesday (W) and so Thursday (Th), Friday (F) and Saturday (S).
The all possible outcomes are :
(T, T), (T, W), (T, Th), (T, F), (T, S)
(W, T), (W, W), (W, Th), (W, F), (W, S)
(Th, T), (Th, W), (Th, Th), (Th, F), (Th, S)
(F, T), (F, W), (F, Th), (F, F), (F, S)
(S, T), (S, W), (S, Tb), (S, F), (S, S)
The total number of all possible outcomes = 25
(i) The favourable outcomes of the visit made by two (RBSESolutions.com) customers on same day = (T, T), (W, W), (Th, Th), (F, F), (S, S)
∴ The number of favourable outcomes = 5.
∴ The probability that the two customers visit the shop on same day = \(\frac { 5 }{ 25 }\) = \(\frac { 1 }{ 5 }\)
⇒ P(A) = \(\frac { 1 }{ 5 }\)
(ii) The favourable outcomes of visiting two customer on two consecutive days = (Shyam, Ekta) = (T, W), (W, Th), (Th, F), (F, S) or (Ekta, Shyam) = (W, T), (Th, W), (F, Th), (S, F)
∴ The number of favourable outcomes = 8
∴ The probability that two customers made the visit to the shop on consecutive days = \(\frac { 8 }{ 25 }\)
(iii) The probability ‘not visit’ made by two customers P(\(\overline { A }\)) = 1 – P(A) = 1 – \(\frac { 1 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
Hence, (i) the probability that two customers visit the shop on same day = \(\frac { 1 }{ 5 }\)
(ii) The probability that two customers visit the shop on two consecutive days = \(\frac { 8 }{ 25 }\)
(iii) The probability that two customers visit the shop on different days = \(\frac { 4 }{ 5 }\)
Question 7.
All the kings, Queens and jacks are taken out from a (RBSESolutions.com) deck of 52 playing cards, then one card is drawn from this well shuffled deck. Find the probability that the card drawn is
(i) A black coloured face card
(ii) A red card. (CBSE 2012)
Solution :
Total number of playing cards in a deck = 52.
The number of face cards i.e., kings, queens, jacks = 12
Taking out all face cards the number of remaining cards = 52 – 12 = 40.
∴ Total number of all possible outcomes = 40
(i) Number of face cards in the remaining playing cards.
Let the event that the black face card drawn be A= 0.
∴ The probability that the card drawn is a black face card P(A) = \(\frac { 0 }{ 40 }\) = 0
This is an impossible event.
(ii) Let occurring the event that the card drawn be (B).
Red cards in the remaining cards = 20.
∴ The number of favourable outcomes of event (B) = 20.
∴ P(B) = \(\frac { 20 }{ 40 }\) = \(\frac { 1 }{ 2 }\)
Hence (i) P(A) = 0 (ii) P(B) = \(\frac { 1 }{ 2 }\)
Question 8.
A box contains 25 notes, ₹ 1 is marked on 19 (RBSESolutions.com) notes of them and ₹ 5 on remaining notes. Another box B contains 50 notes in it ₹ 1 is marked on 45 notes of them and ₹ 13 is marked on the remaining notes. All the noted of two given boxes are collected and put them into another box and mixed them properly.A note is at random drawn from this box. What is the probability that the note drawn has some thing else beside ₹ 1. (NCERT Exemplar Problem)
Solution :
Total number of notes in the boxes A and B = 25 + 50 = 75.
∴ Total number of all possible outcomes = 75
The number of notes on which ₹ 1 is marked = 19 + 45 = 64
Number of notes on which ₹ 1 is not marked = 75 – 64 = 11
Let the event that the note contains some thing else but ₹ 1 be (A).
∴ The number of favourable outcomes of event (A) = 11.
⇒ P(A) = \(\frac { 11 }{ 75 }\)
Hence P(A) = \(\frac { 11 }{ 75 }\)
Question 9.
There are 48 mobile phones in a group out of which 42 are (RBSESolutions.com) good, 3 are some defective and other 3 are largely defected. Varnika will purchase the mobile phone only when it is good, but a businessman will buy the mobile phone only when there is defective largely. A phone is chosen at random from these. What is the probability that the chosen phone is:
(i) Accepted by Varnika.
(ii) Accepted by businessman. (NCERT Exemplar Problem)
Solution :
Total number of mobile phones = 48.
∴ Total number of all possible outcomes = 48
(i) Number of good mobile phones = 42
A mobile phone is chosen at random.
Varnika will buy the mobile phone only when it is good
Let the event that the mobile phone is good be (A).
∴ The number of favourable outcomes for (A) = 42.
⇒ P(A) = \(\frac { 42 }{ 48 }\) = \(\frac { 7 }{ 8 }\)
(ii) The businessman will buy the mobile phone (RBSESolutions.com) only when it is defected largely.
The number of mobile phones in which are not defected largely 42 + 3 = 45.
Let the event that the mobile is not defected largely be (B).
∴ The number of favourable outcomes for B = 45.
⇒ P(B) = \(\frac { 45 }{ 48 }\) = \(\frac { 15 }{ 16 }\)
Hence, P(A) = \(\frac { 7 }{ 8 }\), P(B) = \(\frac { 15 }{ 16 }\).
Question 10.
The entrance fee for a game is ₹ 5. In this game 1 coin is tossed thrice. If Shweta tosses the coin and gets one or two heads, she gets back her fee paid. If she gets three heads, she gets double of her entry fee. Otherwise she losses amount she has paid. When a coin tossed thrice find the probability that : (NCERT Exemplar Problem)
(i) She will loss the entry fee.
(ii) She will get the double of entry fee.
(iii) She will only get back her entry fee.
Solution :
In the game 1 coin is tossed thrice, so all possible outcomes are
HHH, HHT, HTH, THH, THT, HTT, TTH, TTT.
∴ Total number of all possible outcomes = 8.
(i) The entry fee in the game is ₹ 5, if three tails (RBSESolutions.com) are obtained, Shweta will loss her entry fee.
Let the event of getting three heads be (A).
The favourable outcomes of event A = (TTT)
∴ The number of favourable events of (A) = 1.
P(A) = \(\frac { 1 }{ 8 }\)
(ii) If Shweta gets three heads, she will get double of entry fee.
Let the event of occurring 3 heads be (B).
∴ The favourable outcomes of event B = (HHH)
The number of favourable outcomes of (B) = 1
⇒ P(B) = \(\frac { 1 }{ 8 }\)
(iii) If she gets one or two heads, she will (RBSESolutions.com) get back her enty fee.
Let the event ofoccuring one or two heads be (C).
The favourable outcomes of event C = HHT, HTH, THH, THT, HTT, TTH.
∴ The number of favourable outcomes of (C) = 6
⇒ P(C) = \(\frac { 6 }{ 8 }\) = \(\frac { 3 }{ 4 }\)
Hence (i) P(A) = \(\frac { 1 }{ 8 }\)
(ii) P(B) = \(\frac { 1 }{ 8 }\)
(iii) P(C) = \(\frac { 3 }{ 4 }\)
Question 11.
In a game for children, there are 8 triangles in which 3 are blue and remaining red. Also, in this game there are 10 squares, in which 6 are blue and remaining red. One of these figure ¡s at random lost. Find the probabilities of following: (NCERT Exemplar Problem)
(i) The lost figure is a triangle.
(ii) It is a square.
(iii) It is a blue square.
(iv) It is a red triangle.
Solution :
In the given game total number of figures = 8 (triangles) + 10 (squares) = 18
∴ Number of all possible outcomes = 18
(i) Let the event of occurring a triangle be (T).
∴ The number of favourable outcomes of T = 8.
⇒ P(T) = \(\frac { 8 }{ 18 }\) = \(\frac { 4 }{ 9 }\)
(ii) Let the event occurring a square be (S).
∴ The number of favourable outcomes of S = 10
⇒ P(S) = \(\frac { 10 }{ 18 }\) = \(\frac { 5 }{ 9 }\)
(iii) Let the event of occurring the (RBSESolutions.com) blue squares be (B).
∴ There are 6 blue squares.
∴ The number of favourable outcomes of (B) = 6
⇒ P(A) = \(\frac { 6 }{ 18 }\) = \(\frac { 1 }{ 3 }\)
(iv) Let the event occurring red triangles be (P)
The number of red triangles = 8 – 3 = 5
∴ The number of favourable outcomes of (R) = 5
⇒ P(R) = \(\frac { 5 }{ 18 }\)
Hence (i) P(T) = \(\frac { 4 }{ 9 }\)
(ii) P(S) = \(\frac { 5 }{ 9 }\)
(iii) P(B) = \(\frac { 1 }{ 3 }\)
(iv) P(R) = \(\frac { 5 }{ 18 }\)
Question 12.
A dice is thrown once, find the (RBSESolutions.com) probability of:
(i) getting an even number.
(ii) getting a number larger than 3.
(iii)A number between 3 and 6.
Solution :
Tossing the die all possible outcomes = 1, 2, 3, 4, 5, 6
∴ Total number of all possible outcomes =6
(i) The even digits are 2, 4, 6
∴ The number of favourable outcomes = 3
∴ Required probability = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
(ii) The digits more than 3 are 4, 5, 6
∴ The number of favourable outcomes = 3.
∴ Requred probability = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
(iii) The digit between 3 and 6 are 4 and 5.
∴ The number of favourable outcomes = 2
∴ Required probability = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
Question 13.
There are 12 balls ¡n a bag in (RBSESolutions.com) which x are white
(i) What is the probability that if ball drawn at random is white.
(ii) If 6 more white balls are put, (i) the probability of the ball drawn is white becomes double, find the value of x.
Solution :
(i) The number of balls in the bag = 12
∴ The number of all possible outcomes = 12
The number of white balls = x
∴ The number of favourable outcomes = x
∴ Required probability = \(\frac { x }{ 12 }\)
(ii) After putting 6 more white balls in the bag, the total number of ball = 12 + 6 = 18
The number of white balls = (x + 6)
The new probability = \(\frac { x+6 }{ 18 }\)
According to the given problem,
\(\frac { x+6 }{ 18 }\) = \(\frac { x }{ 12 }\) × 2
⇒ \(\frac { x+6 }{ 18 }\) = \(\frac { x }{ 6 }\)
⇒ 6x + 36 = 18x
⇒ 18x – 6x = 36 ⇒ 12x = 6
⇒ x = \(\frac { 36 }{ 12 }\) = 3
Question 14.
There are 30 disc in box, on which the (RBSESolutions.com) numbers 1 to 30 are marked. If a disc is drawn at random from the box. find the probability that the disc drawn has :
(i) a two digit number
(ii) a perfect square
Solution :
Given that the base contains 30 discs.
(i) The two digits number 1 to 30 = 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30
The number of favourable outcomes = 21
The total number of all possible outcomes = 30
The probability that the disc drawn has a two digits number
= \(\frac { 21 }{ 30 }\) = \(\frac { 7 }{ 10 }\)
(ii) The perfect square numbers in 1 to 30 = 1, 4, 9, 16, 25.
The number of favourable outcomes = 5
Total number of all possible outcomes = 30
∴ The probability that the disc drawn contains a perfect square number = \(\frac { 5 }{ 30 }\) = \(\frac { 1 }{ 6 }\)
Question 15.
There are 24 marbles in a jar out of them (RBSESolutions.com) some are green and remaining are blue. If a marble is drawn from the jar at random, the probability that the green marble drawn is \(\frac { 2 }{ 3 }\).
Find the number of blue marbles in the jar.
Solution :
The total number of marbles in the jar = 24
∴ Total number of all possible outcomes = 24.
Let the number of green marbles in jar be x.
∴ The number of favourable outcomes = x.
The probability that the marble drawn ¡s green = \(\frac { x }{ 24 }\)
But we have the probability of drawing a green marble = \(\frac { 2 }{ 3 }\)
\(\frac { x }{ 24 }\) = \(\frac { 2 }{ 3 }\)
3x = 2 × 24
x = \(\frac { 2\times 24 }{ 3 }\)
x = 16.
Thus, the number of green marbles in the jar = 16
The number of blue marbles = 24 – 16 = 8
Hence, the number of blue marbles in the jar = 8
Question 16.
20 ball pens are defective among a heap of 144 ball pens. You (RBSESolutions.com) would like to buy a good pen not bed. The shopkeeper takes out a ball pen at random from these and gives you. What is the probability that:
(i) You will buy that ball pen.
(ii) You will not buy that ball pen.
Solution:
The total number of ball pens in the heap = 144.
The number of defective pens = 20
The number of fine ball pens = 144 – 20 = 124
(i) Let the probability of buying a pen be (A).
∵ We would like to buy a good pen
∴ The number of favourable outcomes of buying a ball pen = 124
Total number of all possible outcomes = 144
∴ P(A) = \(\frac { 124 }{ 144 }\) = \(\frac { 31 }{ 36 }\)
(ii) Let the event of not buying a ball pen be (\(\overline { A }\))
Then P(\(\overline { A }\)) = 1 – P(A)
= 1 – \(\frac { 31 }{ 36 }\) = \(\frac { 5 }{ 36 }\)
Hence (i) P(A) = \(\frac { 31 }{ 36 }\)
(ii) P(\(\overline { x }\)) = \(\frac { 5 }{ 36 }\)
Question 17.
(i) Among 20 bulbs, 4 bulbs are defective. A bulb is drawn at (RBSESolutions.com) random from these. What is the probability that the bulb drawn will be defective?
(ii) Let the bulb drawn is neither defective nor it is mixed with the remaining bulbs again. Now a bulb is taken out from these remaining bulbs. What is the probability of the bulb drawn ¡s not defective?
Solution :
(i) The number of total bulbs = 20
The number of defective bulbs = 4
If a bulb is drawn at random, then the number of favourable outcomes of defective bulbs = 4
Total number of all possible outcomes = 20
∴ The required probability = \(\frac { 4 }{ 20 }\) = \(\frac { 1 }{ 5 }\)
(ii) If the bulb drawn in neither defective nor mixed into remaining bulbs.
Again another bulb is taken out.
∴ The total number of all possible outcomes = 20 – 1 = 19
The number of favourable outcomes of the event that the bulb drawn is defective = 4
∴ The probability that the bulb drawn is defective = \(\frac { 4 }{ 19 }\).
∴ The probability that the bulb drawn is not defective = 1 – \(\frac { 4 }{ 19 }\) = \(\frac { 15 }{ 19 }\)
Hence, (i) The probability that the bulb is defective = \(\frac { 1 }{ 5 }\)
(ii) The probability the bulb is not defective = \(\frac { 15 }{ 19 }\)
Question 18.
In a Piggy bank there are 100 coins of 50 paise, 50 (RBSESolutions.com) coins of ₹ 1, 20 coins of ₹ 2 and 10 coins of 5. If after shaking the bank properly and turning it downward the probability of falling a coin is at random. Find the probability that coin fallen is:
(i) 50 paise
(ii) Not ₹ 5.
Solution :
Number of 50 paise coins = 100
Number of ₹ 1 coins 50
Number of ₹ 2 coins 20
Number of ₹ 5 coins = 10
Total number of coins = 100 + 50 + 20 + 10 = 180
∴ Total number of all possible outcomes = 180
(i) The bank contains 100 coins of 50 paise.
∴ The number of favourable outcomes = 100
∴ The probability of the coin fallen is a coin of 50 paise
∴ P(50 paise coin) = \(\frac { 5 }{ 9 }\)
(ii) The number of ₹ 5 coins = 10
∴ Probability of ₹ 5
= \(\frac { 10 }{ 180 }\) = p(₹ 5 coin) = \(\frac { 1 }{ 18 }\)
The probability that the coin fallen is not ₹ 5.
P(not a coin of ₹ 5) = 1 – P(₹ 5 coin)
= 1 – \(\frac { 1 }{ 18 }\) = \(\frac { 17 }{ 18 }\)
Hence (i) P (50 paise coin) = \(\frac { 5 }{ 9 }\)
(ii) p(not ₹ 5 coin) = \(\frac { 17 }{ 18 }\)
Question 19.
A bag contains 24 balls In which red balls x, white balls 2x and blue balls 3x. A ball at (RBSESolutions.com) random drawn. What ¡s the probability that the ball drawn is :
(i) Not red (ii) White (NCERT Exemplar Problem)
Solution :
The number of red, white and blue balls in the bag are x, 2x and 3x respectively.
Total number of balls = 24
⇒ x + 2x + 3x = 24
6x = 24
x = \(\frac { 24 }{ 6 }\) = 4
∴ Red balls = 4
White balls = 2 × 4 = 8
And blue balls = 3 × 4 = 12
Total number of all possible outcomes = 24.
(i) Number of balls which are not red = 8 + 12 = 20
∴ The favourable outcomes of not a red ball = 20
∴ The probability that the ball drawn is not red
= \(\frac { 20 }{ 24 }\) = \(\frac { 5 }{ 6 }\)
(ii) The number of favourable outcomes of the ball (RBSESolutions.com) drawn is white = 8.
∴ The probability that the drawn ball is white = \(\frac { 8 }{ 24 }\) = \(\frac { 1 }{ 3 }\)
Hence, (i) the probability that the ball not red = \(\frac { 5 }{ 6 }\)
(ii) the probability that the ball is white = \(\frac { 1 }{ 3 }\)
Question 20.
On the 6 faces of a dice, the numbers 0, 1, 1, 1, 6, 6 are marked. Such these both dice are thrown at a time and the sum of the digits on the top faces is noted.
(i) How many total number of possible outcomes.
(ii) What is the probability of getting a sum of 7 on the top faces.
Solution :
Two dices contain the numbers 0, 1, 1, 1, 6, 6 on the faces each.
The all possible outcomes will be:
(0, 0), (0, 1), (0, 1), (0, 1), (0, 6), (0, 6)
(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(1,0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)
(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)
(i) The possible numbers which have different outcomes are (0, 0), (0, 1), (0, 6), (1, 0), (1, 1), (1, 6), (6, 6)
Hence, the possible outcomes are : (0, 1, 2, 6, 7, 12)
(ii) Total number of all possible outcomes = 36
The numbers that have a sum 7 are
(1,6), (1,6), (1,6), (1,6), (1,6), (1,6)
(6, 1), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1)
The number of favourable outcomes = 12
∴ The probability of getting a sum of 7 = \(\frac { 12 }{ 36 }\) = \(\frac { 1 }{ 3 }\)
Hence (i) All possible different outcomes = (0, 1, 2, 6, 7, 12)
(ii) The probability of getting a sum 7 = \(\frac { 1 }{ 3 }\)
Question 21.
A dice is thrown twice. What is the probability if:
(i) 5 will be never occured
(ii) less than 5 will be occure once.
Solution :
When a dice is thrown twice, the all (RBSESolutions.com) possible outcomes are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2 ,4), (2, 5), (2, 6)
(3, 1), (3, 2,), (3, 3), (3, 4), (3, 5), (3 ,6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total number of all possible outcomes = 36
Number of the outcomes which contain digit 5 = 11.
The number which does not contain the digit 5 = 36 – 11 = 25
(i) Number of favourable outcomes not occurring 5 = 25
Total number of all possible outcomes = 36
The required probability
= \(\frac { 25 }{ 36 }\)
(ii) The number of favourable outcomes of (RBSESolutions.com) occurring a digit 5 at least = 11
Total number of all possible outcomes = 36
∴ The probability that the digit 5 occurs at least one time = \(\frac { 11 }{ 36 }\)
Hence (i) The probability not occurring 5 = \(\frac { 25 }{ 36 }\)
(ii) Probability of occurring 5 at least one time = \(\frac { 11 }{ 36 }\)
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