This exercise deals with the axiomatic approach to the probability that is deduced from mathematical concepts. It is based upon certain axioms. The main aim of subject experts is to provide solutions and thereby help students, irrespective of their intelligence quotient. The solutions are explained and solved in simple language to help them perform well in the board exams. Students are advised to solve exercise-wise problems on a daily basis to understand the concepts in a better way. Students can download RD Sharma Class 11 Maths Solutions from the links given below and can use it for future reference as well.
RD Sharma Solutions for Class 11 Maths Exercise 33.3 Chapter 33 – Probability
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EXERCISE 33.3 PAGE NO: 33.45
1. Which of the following cannot be a valid assignment of probability for elementary events or outcomes of sample space S = {w1, w2, w3, w4, w5, w6, w7}:
Solution:
For each event to be a valid assignment of probability.
The probability of each event in the sample space should be less than 1, and the sum of the probability of all the events should be exactly equal to 1.
(i) It is valid as each P (wi) (for i=1 to 7) lies between 0 to 1, and the sum of P (w1) =1
(ii) It is valid as each P (wi) (for i=1 to 7) lies between 0 to 1, and the sum of P (w1) =1
(iii) It is not valid as the sum of P (wi) = 2.8, which is greater than 1
(iv) it is not valid as P (w7) = 15/14, which is greater than 1
2. A die is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
Solution:
Given: A die is thrown.
The total number of outcomes is six, n (S) = 6
By using the formula,
P (E) = favourable outcomes / total possible outcomes
(i) Let E be the event of getting a prime number.
E = {2, 3, 5}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 6
= ½
(ii) Let E be the event of getting 2 or 4.
E = {2, 4}
n (E) = 2
P (E) = n (E) / n (S)
= 2 / 6
= 1/3
(iii) Let E be the event of getting a multiple of 2 or 3.
E = {2, 3, 4, 6}
n (E) = 4
P (E) = n (E) / n (S)
= 4 / 6
= 2/3
3. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) an even number on first
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 is the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) neither a doublet nor a total of 10
(xiii) odd number on the first and 6 on the second
(xiv) a number greater than 4 on each die
(xv) a total of 9 or 11
(xvi) a total greater than 8
Solution:
Given: A pair of dice has been thrown, so the number of elementary events in sample space is 62 = 36
n (S) = 36
By using the formula,
P (E) = favourable outcomes / total possible outcomes
(i) Let E be the event that the sum 8 appears.
E = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n (E) = 5
P (E) = n (E) / n (S)
= 5 / 36
(ii) Let E be the event of getting a doublet.
E = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(iii) Let E be the event of getting a doublet of prime numbers.
E = {((2, 2) (3, 3) (5, 5)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(iv) Let E be the event of getting a doublet of odd numbers.
E = {(1, 1) (3, 3) (5, 5)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(v) Let E be the event of getting a sum greater than 9.
E = {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(vi) Let E be the event of getting even on the first die.
E = {(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
n (E) = 18
P (E) = n (E) / n (S)
= 18 / 36
= ½
(vii) Let E be the event of getting even on one and multiple of three on the other.
E = {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}
n (E) = 11
P (E) = n (E) / n (S)
= 11 / 36
(viii) Let E be the event of getting neither 9 nor 11 as the sum.
E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(ix) Let E be the event of getting a sum less than 6.
E = {(1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) (3,1) (3,2) (4,1)}
n (E) = 10
P (E) = n (E) / n (S)
= 10 / 36
= 5/18
(x) Let E be the event of getting a sum less than 7.
E = {(1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2) (2,3) (2,4) (3,1) (3,2) (3,3) (4,1) (4,2) (5,1)}
n (E) = 15
P (E) = n (E) / n (S)
= 15 / 36
= 5/12
(xi) Let E be the event of getting more than 7.
E = {(2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5) (5,6) (6,2) (6,3) (6,4) (6,5) (6,6)}
n (E) = 15
P (E) = n (E) / n (S)
= 15 / 36
= 5/12
(xii) Let E be the event of getting neither a doublet nor a total of 10.
E′ be the event that either a double or a sum of ten appears.
E′ = {(1,1) (2,2) (3,3) (4,6) (5,5) (6,4) (6,6) (4,4)}
n (E′) = 8
P (E′) = n (E′) / n (S)
= 8 / 36
= 2/9
So, P (E) = 1 – P (E′)
= 1 – 2/9
= 7/9
(xiii) Let E be the event of getting an odd number on the first and 6 on the second.
E = {(1,6) (5,6) (3,6)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(xiv) Let E be the event of getting greater than 4 on each die.
E = {(5,5) (5,6) (6,5) (6,6)}
n (E) = 4
P (E) = n (E) / n (S)
= 4 / 36
= 1/9
(xv) Let E be the event of getting a total of 9 or 11.
E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(xvi) Let E be the event of getting a total greater than 8.
E = {(3,6) (4,5) (4,6) (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6)}
n (E) = 10
P (E) = n (E) / n (S)
= 10 / 36
= 5/18
4. In a single throw of three dice, find the probability of getting a total of 17 or 18
Solution:
Given: The dice are thrown.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Total number of possible outcomes is 63=216
So, n (S) = 216
Let E be the event of getting a total of 17 or 18.
E = {(6, 6, 5) (6, 5, 6) (5, 6, 6) (6, 6, 6)}
n (E) = 4
P (E) = n (E) / n (S)
= 4 / 216
= 1/54
5. Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail
Solution:
Given: Three coins are tossed together.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Total number of possible outcomes is 23 = 8
(i) Let E be the event of getting exactly two heads.
E = {(H, H, T) (H, T, H) (T, H, H)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 8
(ii) Let E be the event of getting at least two heads.
E= {(H, H, T) (H, T, H) (T, H, H) (H, H, H)}
n (E)=4
P (E) = n (E) / n (S)
= 4 / 8
= ½
(iii) Let E be the event of getting at least one head and one tail.
E = {(H, T, T) (T, H, T) (T, T, H) (H, H, T) (H, T, H) (T, H, H)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 8
= ¾
6. What is the probability that an ordinary year has 53 Sundays?
Solution:
Given: A year which includes 52 weeks and one day.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
So, we now have to determine the probability of that one day being Sunday.
Total number of possible outcomes is 7.
n(S) = 7
E = {M, T, W, T, F, S, SU}
n (E) = 1
P (E) = n (E) / n (S)
= 1 / 7
7. What is the probability that a leap year has 53 Sundays and 53 Mondays?
Solution:
Given: A leap year which includes 52 weeks and two days.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
So, we now have to determine the probability of the remaining two days being Sunday and Monday.
S = {MT, TW, WT, TF, FS, SSu, SuM}
n (S) = 7
E= {SuM}
n (E) = 1
P (E) = n (E) / n (S)
= 1 / 7
8. A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:
(i) All three balls are white.
(ii) All three balls are red.
(iii) One ball is red, and two balls are white.
Solution:
Given: A bag contains 8 red and 5 white balls.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Total number of ways of drawing three balls at random is 13C3
n (S) = 286
(i) Let E be the event of getting all white balls.
E= {(W) (W) (W)}
n (E)= 5C3=10
P (E) = n (E) / n (S)
= 10 / 286
= 5/143
(ii) Let E be the event of getting all red balls.
E = {(R) (R) (R)}
n (E)= 8C3 =56
P (E) = n (E) / n (S)
= 56 / 286
= 28/143
(iii) Let E be the event of getting one red and two white balls.
E = {(R…. 80th R)}
n (E)= 8C15C2 = 80
P (E) = n (E) / n (S)
= 80 / 286
= 40/143
9. In a single throw of three dice, find the probability of getting the same number on all the three dice
Solution:
Given: Three dice are rolled over.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
So, we now have to determine the probability of getting the same number on all three dice.
Total number of possible outcomes is 63=216
n (S) = 216
Let E be the event of getting the same number on all the three dice.
E = {(1,1,1) (2,2,2) (3,3,3) (4,4,4) (5,5,5) (6,6,6)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 216
= 1/36
10. Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10
Solution:
Given: Two unbiased dice are thrown.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
So, we now have to determine the probability of getting the sum of digits on dice greater than 10.
Total number of possible outcomes is 62=36
n (S) = 36
Let E be the event of getting the same number on all three dice.
E = {(5,6) (6,5) (6,6)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
11. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither an ace nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) a diamond card
(ix) not a diamond card
(x) a black card
(xi) not an ace
(xii) not a black card
Solution:
Given: Pack of 52 cards.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
We know that a card is drawn from a pack of 52 cards, so the number of elementary events in the sample space is
n (S) = 52C1 = 52
(i) Let E be the event of drawing a black king
n (E) =2C1 =2 (there are two black kings, one of spade and the other of the club)
P (E) = n (E) / n (S)
= 2 / 52
= 1/26
(ii) Let E be the event of drawing a black card or a king
n (E) = 26C1+4C1–2C1= 28
[We are subtracting 2 from the total because there are two black kings, which are already counted, and to avoid the error of considering it twice.]P (E) = n (E) / n (S)
= 28 / 52
= 7/13
(iii) Let E be the event of drawing a black card and a king.
n (E) =2C1 = 2 (there are two black kings, one of spade and the other of the club.)
P (E) = n (E) / n (S)
= 2 / 52
= 1/26
(iv) Let E be the event of drawing a jack, queen or king.
n (E) = 4C1+4C1+4C1 = 12
P (E) = n (E) / n (S)
= 12 / 52
= 3/13
(v) Let E be the event of drawing neither a heart nor a king.
Now, let us consider E′ as the event that either a heart or king appears,
n (E′) = 6C1+4C1-1=16 (there is a heart king, so it is deducted.)
P (E′) = n (E′) / n (S)
= 16 / 52
= 4/13
So, P (E) = 1 – P (E′)
= 1 – 4/13
= 9/13
(vi) Let E be the event of drawing a spade or king.
n (E)=13C1+4C1-1=16
P (E) = n (E) / n (S)
= 16 / 52
= 4/13
(vii) Let E be the event of drawing neither an ace nor a king.
Now, let us consider E′ as the event that either an ace or king appears.
n(E′) = 4C1+4C1 = 8
P (E′) = n (E′) / n (S)
= 8 / 52
= 2/13
So, P (E) = 1 – P (E′)
= 1 – 2/13
= 11/13
(viii) Let E be the event of drawing a diamond card.
n (E)=13C1=13
P (E) = n (E) / n (S)
= 13 / 52
= ¼
(ix) Let E be the event of drawing, not a diamond card.
Now, let us consider E′ as the event that a diamond card appears.
n (E′) =13C1=13
P (E′) = n (E′) / n (S)
= 13 / 52
= 1/4
So, P (E) = 1 – P (E′)
= 1 – 1/4
= ¾
(x) Let E be the event of drawing a black card.
n (E) =26C1 = 26 (spades and clubs)
P (E) = n (E) / n (S)
= 26 / 52
= ½
(xi) Let E be the event of drawing, not an ace.
Now, let us consider E′ as the event that an ace card appears.
n (E′) = 4C1 = 4
P (E′) = n (E′) / n (S)
= 4 / 52
= 1/13
So, P (E) = 1 – P (E′)
= 1 – 1/13
=12/13
(xii) Let E be the event of not drawing a black card.
n (E) = 26C1 = 26 (red cards of hearts and diamonds)
P (E) = n (E) / n (S)
= 26 / 52
= ½
12. In shutting a pack of 52 playing cards, four are accidentally dropped; find the chance that the missing cards should be one from each suit.
Solution:
Given: A pack of 52 cards from which 4 are dropped.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
We now have to find the probability that the missing cards should be one from each suit.
We know that, from a well-shuffled pack of cards, 4 cards missed out on total possible outcomes are
n (S) = 52C4 = 270725
Let E be the event that four missing cards are from each suite.
n (E) = 13C1×13C1×13C1×13C1 = 134
P (E) = n (E) / n (S)
= 134 / 270725
= 2197/20825
13. From a deck of 52 cards, four cards are drawn simultaneously, find the chance that they will be the four honours of the same suit.
Solution:
Given: A deck of 52 cards.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
We now have to find the probability that all the face cards of the same suits are drawn.
Total possible outcomes are
n (S) = 52C4
Let E be the event that all the cards drawn are face cards of the same suit.
n (E)=4×4C4=4
P (E) = n (E) / n (S)
= 4 / 270725
14. Tickets numbered from 1 to 20 are mixed up together, and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?
Solution:
Given: Numbered tickets from 1 to 20.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
To find the probability of the ticket drawn having a number which is a multiple of 3 or 7.
We know that one ticket is drawn from a lot of mixed numbers.
Total possible outcomes are
n (S) = 20C1 = 20
Let E be the event of getting a ticket which has a number that is multiple of 3 or 7.
E = {3,6,9,12,15,18,7,14}
n (E) = 8
P (E) = n (E) / n (S)
= 8 / 20
= 2/5
15. A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that one is red, one is white, and one is blue.
Solution:
Given: A bag containing 6 red, 4 white and 8 blue balls.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Three balls are drawn, so we have to find the probability that one is red, one is white, and one is blue.
Total number of outcomes for drawing 3 balls is 18C3
n (S) = 18C3 = 816
Let E be the event that one red, one white and one blue ball is drawn.
n (E) = 6C14C18C1 = 192
P (E) = n (E) / n (S)
= 192 / 816
= 4/17
16. A bag contains 7 white, 5 black and 4 red balls. If two balls are drawn at random, find the probability that:
(i) both the balls are white
(ii) one ball is black, and the other red
(iii) both the balls are of the same colour
Solution:
Given: A bag containing 7 white, 5 black and 4 red balls.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Two balls are drawn at random, therefore
Total possible outcomes are 16C2
n (S) = 120
(i) Let E be the event of getting both white balls.
E = {(W) (W)}
n (E) = 7C2 = 21
P (E) = n (E) / n (S)
= 21 / 120
= 7/40
(ii) Let E be the event of getting one black and one red ball.
E = {(B) (R)}
n (E) = 5C14C1 = 20
P (E) = n (E) / n (S)
= 20 / 120
= 1/6
(iii) Let E be the event of getting both balls of the same colour.
E = {(B) (B)} or {(W) (W)} or {(R) (R)}
n (E) = 7C2+5C2+4C2 = 37
P (E) = n (E) / n (S)
= 37 / 120
17. A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that:
(i) one is red, and two are white
(ii) two are blue, and one is red
(iii) one is red
Solution:
Given: A bag containing 6 red, 4 white and 8 blue balls.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Two balls are drawn at random.
Total possible outcomes are 18C3
n (S) = 816
(i) Let E be the event of getting one red and two white balls.
E = {(W) (W) (R)}
n (E) = 6C14C2 = 36
P (E) = n (E) / n (S)
= 36 / 816
= 3/68
(ii) Let E be the event of getting two blue and one red.
E = {(B) (B) (R)}
n (E) = 8C26C1 = 168
P (E) = n (E) / n (S)
= 168 / 816
= 7/34
(iii) Let E be the event that one of the balls must be red.
E = {(R) (B) (B)} or {(R) (W) (W)} or {(R) (B) (W)}
n (E) = 6C14C18C1+6C14C2+6C18C2 = 396
P (E) = n (E) / n (S)
= 396 / 816
= 33/68
18. Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain:
(i) just one ace
(ii) at least one ace?
Solution:
Given: Five cards are drawn from a pack of 52 cards.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Five cards are drawn at random.
Total possible outcomes are 52C5
n (S) = 2598960
(i) Let E be the event that exactly only one ace is present.
n (E) = 4C148C4 = 778320
P (E) = n (E) / n (S)
= 778320 / 2598960
= 3243/10829
(ii) Let E be the event that at least one ace is present.
E = {1 or 2 or 3 or 4 ace(s)}
n (E) = 4C148C4+4C248C3+4C348C2+4C448C1 = 886656
P (E) = n (E) / n (S)
= 886656 / 2598960
= 18472/54145
19. The face cards are removed from a full pack. Out of the remaining 40 cards, 4 are drawn at random. What is the probability that they belong to different suits?
Solution:
Given: The face cards are removed from a full pack of 52.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Four cards are drawn from the remaining 40 cards, so we have to find the probability that all of them belong to a different suit.
Total possible outcomes of drawing four cards are 40C4
n (S) = 40C4 = 91390
Let E be the event that 4 cards belong to a different suit.
n (E) = 10C110C110C110C1 = 10000
P (E) = n (E) / n (S)
= 10000 / 91390
= 1000/9139
20. There are four men and six women on the city councils. If one council member is selected for a committee at random, how likely is it that it is a woman?
Solution:
Given: There are four men and six women on the city councils.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
From the city council, one person is selected as a council member, so we have to find the probability that it is a woman.
Total possible outcomes of selecting a person are 10C1
n (S)= 10C1 = 10
Let E be the event that it is a woman.
n (E) = 6C1 = 6
P (E) = n (E) / n (S)
= 6 / 10
= 3/5
21. A box contains 100bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that:
(i) all 10 are defective
(ii) all 10 are good
(iii) at least one is defective
(iv) none is defective
Solution:
Given: A box contains 100 bulbs, 20 of which are defective.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
Ten bulbs are drawn at random for inspection.
Total possible outcomes are 100C10
n (S) = 100C10
(i) Let E be the event that all ten bulbs are defective.
n (E) = 20C10
P (E) = n (E) / n (S)
= 20C10 / 100C10
(ii) Let E be the event that all ten good bulbs are selected.
n (E) = 80C10
P (E) = n (E) / n (S)
= 80C10 / 100C10
(iii) Let E be the event that at least one bulb is defective.
E= {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs.
Let E′ be the event that none of the bulbs is defective.
n (E′) = 80C10
P (E′) = n (E′) / n (S)
= 80C10 / 100C10
So, P (E) = 1 – P (E′)
= 1 – 80C10 / 100C10
(iv) Let E be the event that none of the selected bulbs is defective.
n (E) = 80C10
P (E) = n (E) / n (S)
= 80C10 / 100C10
22. Find the probability that in a random arrangement of the letters of the word ‘SOCIAL’ vowels come together.
Solution:
Given: The word ‘SOCIAL’.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
In the random arrangement of the alphabet of the word “SOCIAL,” we have to find the probability that vowels come together.
Total possible outcomes of arranging the alphabet are 6!
n (S) = 6!
Let E be the event that vowels come together.
The vowels in SOCIAL are A, I, O.
So, the number of ways to arrange them where (A, I, O) come together is
n (E) = 4! × 3!
P (E) = n (E) / n (S)
= [4! × 3!] / 6!
= 1/5
23. The letters of the word ‘CLIFTON’ are placed at random in a row. What is the chance that two vowels come together?
Solution:
Given: The word ‘CLIFTON’.
By using the formula,
P (E) = favourable outcomes / total possible outcomes
In the random arrangement of the alphabet of the word “CLIFTON”, we have to find the probability that vowels come together.
Total possible outcomes of arranging the alphabet are 7!
n (S) =7!
Let E be the event that vowels come together.
The vowels in CLIFTON are I, O.
The number of ways to arrange them where (I, O) comes together.
n (E)= 6! × 2!
P (E) = n (E) / n (S)
= [6! × 2!] / 7!
= 2/7
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