Question 1: Which of the following cannot be valid assignment of probability for elementary events or
outcomes of sample space $S = \{ w_1, w_2, w_3, w_4, w_5, w_6, w_7 \}$

 Elementary Events $w_1$$w_1$ $w_2$$w_2$ $w_3$$w_3$ $w_4$$w_4$ $w_5$$w_5$ $w_6$$w_6$ $w_7$$w_7$ (i) $0.1$$0.1$ $0.01$$0.01$ $0.05$$0.05$ $0.03$$0.03$ $0.01$$0.01$ $0.2$$0.2$ $0.6$$0.6$ (ii) $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ (iii) $0.7$$0.7$ $0.6$$0.6$ $0.5$$0.5$ $0.4$$0.4$ $0.3$$0.3$ $0.2$$0.2$ $0.1$$0.1$ (iv) $\displaystyle \frac{1}{14}$$\displaystyle \frac{1}{14}$ $\displaystyle \frac{2}{14}$$\displaystyle \frac{2}{14}$ $\displaystyle \frac{3}{14}$$\displaystyle \frac{3}{14}$ $\displaystyle \frac{4}{14}$$\displaystyle \frac{4}{14}$ $\displaystyle \frac{5}{14}$$\displaystyle \frac{5}{14}$ $\displaystyle \frac{6}{14}$$\displaystyle \frac{6}{14}$ $\displaystyle \frac{15}{14}$$\displaystyle \frac{15}{14}$

For each event to be a valid assignment of probability, the probability of each event in sample space should be less than 1 and the sum of probability of all the events should be exactly equal to 1

(i) it is valid as each $P(w_i) ( \ for \ i=1 \ to \ 7)$ lies between 0 to 1 and sum of $P(w_1) = 1$

(ii) it is valid as each $P(w_i) ( \ for \ i = 1 \ to \ 7)$ lies between 0 to 1 and sum of $P(w_1) =1$

(iii) it is not valid as sum of $P(w_i)=2.8$ which is greater than 1

(iv) it is not valid as $\displaystyle P(w_7) = \frac{15 }{14 }$ which is greater than 1.

$\\$

Question 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.

(iv) an even prime number          (v) a number greater than 5

(vi) a number lying between 2 and 6

(i) Prime numbers on a dice are 2, 3, and 5. So, the total number of prime numbers is 3. We know that, Probability = Number of favorable outcomes/ Total number of outcomes Thus, probability of getting a prime number $\displaystyle = \frac{3 }{6} = \frac{1}{2}$

(ii) For getting 2 and 4, clearly the number of favorable outcomes is 2. We know that Probability = Number of favorable outcomes/ Total number of  $\displaystyle \text{outcomes Thus, the probability of getting 2 or 4} = \frac{2 }{6} = \frac{1}{3}$

(iii) Multiple of 2 are 3 are 2, 3, 4 and 6. So, the number of favorable outcomes is 4 We know that, Probability = Number of favorable outcomes/ Total number  $\displaystyle \text{of outcomes Thus, the probability of getting an multiple of 2 or 3} = \frac{4 }{6} = \frac{2}{3}$

(iv) An even prime number is 2 only. So, the number of favorable outcomes is 1. We know that, Probability = Number of favorable outcomes/ Total number of $\displaystyle \text{outcomes Thus, the probability of getting an even prime number }= \frac{1}{6}$

(v) A number greater than 5 is 6 only. So, the number of favorable outcomes is 1. We know that, Probability = Number of favorable outcomes/ Total number of  $\displaystyle \text{outcomes Thus, the probability of getting a number greater than 5} = \frac{1}{6}$

(vi) Total number on a dice is 6. Numbers lying between 2 and 6 are 3, 4 and 5 So, the total number of numbers lying between 2 and 6 is 3. We know that, Probability = Number of favorable outcomes/ Total number of outcomes Thus, $\displaystyle \text{the probability of getting a number lying between 2 and 6 } = \frac{3}{6} = \frac{1}{2}$

$\\$

Question 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) a doublet of odd numbers     (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 as 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

In a throw of pair of dice, total no of possible outcomes $= 36 (6 \times 6)$ which are

$(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3) ,(2, 4), (2, 5), (2, 6)$

$(3, 1), (3, 2) ,(3, 3), (3, 4) ,(3, 5), (3, 6), (4, 1), (4, 2), (4, 3) ,(4, 4), (4, 5), (4, 6)$

$(5, 1), (5, 2), (5, 3) ,(5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$

(i) Let E be event of getting the sum as 8

No. of favorable outcomes $= 5 \text{ i.e. } \{ (2, 6) , (3, 5) , (4, 4) , (5, 3) , (6, 2) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{5}{36}$

(ii) Let E be the event of getting a doublet

No. of favorable outcomes $= 5 \text{ i.e. } \{ (1, 1) , (2, 2) , (3, 3) , (4, 4) , (5, 5) , (6, 6) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{6}{36} = \frac{1}{6}$

(iii) Let E be the event of getting a doublet of prime no’s

No. of favorable outcomes $= 3 \text{ i.e. } \{ (2, 2) , (3, 3) , (5, 5) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{3}{36} = \frac{1}{12}$

(iv) Let E be the event of getting a doublet of odd no’s

No. of favorable outcomes $= 3 \text{ i.e. } \{ (1, 1) , (3, 3) , (5, 5) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{3}{36} = \frac{1}{12}$

(v) E ⟶ event of getting a sum greater than 9

No. of favorable outcomes $= 6 \text{ i.e. } \{ (4, 6) , (5, 5) , (5, 6) , (6, 4) , (6, 5) , (6, 6) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{6}{36} = \frac{1}{6}$

(vi) Let E be the event of getting an even no. on first

No. of favorable outcomes $= 18 \text{ i.e. } \{(2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 5) , (2, 6) , (4, 1) , (4, 2) , (4, 3) , \\ (4, 4) , (4, 5) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{18}{36} = \frac{1}{3}$

(vii) Let E be the event of getting an even no. on one and a multiple of 3 on other

No. of favorable outcomes $= 11 \text{ i.e. } \{ (2, 3) , (2, 6) , (4, 3) , (4, 6) , (6, 3) , (6, 6) , (3, 2) , (3, 4) , (3, 4) , (3, 6) , (6, 2) , (6, 4) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{11}{36}$

(viii) Let $\overline{E}$ be the event of getting neither 9 nor 11 as the sum of numbers on faces

Therefore E is  getting either 9 or 11 as the sum of no’s on faces

No. of favorable outcomes $= 6 \text{ i.e. } \{ (3, 6) , (4, 5) , (5, 4) , (6, 3) , (5, 6) , (6, 5) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{6}{36} = \frac{1}{6}$

$\displaystyle P(\overline{E}) = 1 - P(E) = 1 - \frac{1}{6} = \frac{5}{6}$

(ix) Let E be the event of getting a sum less than 6

No. of favorable outcomes $= 10 \text{ i.e. } \{(1, 1) , (1, 2) , (1, 3) , (1, 4) , (2, 1) , (2, 2) , (2, 3) , (3, 1) , (3, 2) , (4, 1)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{10}{36} = \frac{5}{18}$

(x) Let E be the event of getting a sum less than 7

No. of favorable outcomes $= 15 \text{ i.e. } \{(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , \\ (3, 1) , (3, 2) , (3, 3) , (4, 1) , (4, 2) , (5, 1)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{15}{36} = \frac{5}{12}$

(xi) Let E be the event of getting a sum more than 7

No. of favorable outcomes $= 15 \text{ i.e. } \{(2, 6) , (3, 5) , (3, 6) , (4, 4) , (4, 5) , (4, 6) , (5, 3) , (5, 4) , (5, 5) , \\ (5, 6) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{15}{36} = \frac{5}{12}$

(xii) Let E be the event of getting a 1 at least once

No. of favorable outcomes $= 11 \text{ i.e. } \{(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (1, 6) , (2, 1) , (3, 1) , (4, 1) , (5, 1) , (6, 1)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{11}{36}$

(xiii) Let E be the event of getting a no other than 5 on any dice

No. of favorable outcomes $= 25 \text{ i.e. } \{(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 6) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 6) , (3, 1) , \\ (3, 2) , (3, 3) , (3, 4) , (3, 6) , (4, 1) , (4, 2) , (4, 3) , (4, 4) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 6)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{25}{36}$

$\\$

Question 4: In a single throw of three dice, find the probability of getting a total of 17 or 18.

Three dices are thrown once.

Total number of possible outcomes is $\displaystyle 6^3 = 216$

$\displaystyle \text{Therefore, } n (S) = 216$

Let E be the event of getting total of 17 or 18

$\displaystyle \text{Therefore } E = 4 \text{ i.e.} \{ (6, 6, 5), (6, 5, 6), (5, 6, 6), (6, 6, 6) \}$

$\displaystyle n (E) = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 216} = \frac{1}{54}$

$\\$

Question 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.

Three coins are tossed together

Total number of possible outcomes is $\displaystyle 2^3 = 8$

$\displaystyle \text{Therefore, } n (S) = 8$

Sample space for three coins tossed is  $\displaystyle \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}$

(i) exactly two heads

$\displaystyle n (E) = 3$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{3 }{ 8}$

(ii) at least two heads

$\displaystyle n (E) = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 8} = \frac{1}{2}$

(iii) at least one head and one tail.

$\displaystyle n (E) = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 8} = \frac{1}{4}$

$\\$

Question 6: What is the probability that an ordinary year has 53 Sundays?

Ordinary year has 365 days

365 days = 52 weeks + 1 day

That 1 day may be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

Total no. of possible outcomes $\displaystyle n(S) = 7$

Let E be the event of getting 53 Sundays

No. of favorable outcomes = 1 {Sunday}

$\displaystyle \text{No. of favorable outcomes} n (E) = 1$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1 }{ 7}$

$\\$

Question 7: What is the probability that a leap year has 53 Sundays and 53 Mondays?

A leap year consists of 366 days comprising of 52 weeks and 2 days.

There are 7 possibilities for these 2 extra days viz.

(i) Sunday, Monday     (ii) Monday, Tuesday     (iii) Tuesday, Wednesday,

(iv) Wednesday, Thursday     (v) Thursday, Friday     (vi) Friday, Saturday and

(vii) Saturday, Sunday.

Let us consider two events :

A : the leap year contains 53 Sundays

B : the leap year contains 53 Mondays.

$\displaystyle \text{Then we have } P(A) = \frac{2}{7}, \ \ \ P(B) = \frac{2}{7}$

$\displaystyle P(A \cap B) = \frac{1}{7}$

Hence, required probability $\displaystyle = P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}$

$\\$

Question 8: A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that: (i) All the three balls are white.      (ii) A11 the three balls are red.        (iii) One ball is red and two balls are white

A bag contains 8 red and 5 white balls.

$\displaystyle \text{ Total number of possible outcomes } n(S) = ^{13}C_{3}$

(i) Probability when all the three balls are white

$\displaystyle n(E) = ^{5}C_{3}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{5}C_{3} }{ ^{13}C_{3}} = \frac{ \frac{5!}{3! 2!} }{ \frac{13!}{3! 10!} } = \frac{10}{ \frac{13 \times 12 \times 11}{6} } = \frac{5}{143}$

(ii) Probability when all the three balls are red

$\displaystyle n(E) = ^{8}C_{3}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{8}C_{3} }{ ^{13}C_{3}} = \frac{ \frac{8!}{5! 3!} }{ \frac{13!}{3! 10!} } = \frac{ \frac{8 \times 7 \times 6}{6} }{ \frac{13 \times 12 \times 11}{6} } = \frac{28}{143}$

(iii) Probability when One ball is red and two balls are white

$\displaystyle n(E) = ^{8}C_{1} \times ^{5}C_{2}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{8}C_{1} \times ^{5}C_{2} }{ ^{13}C_{3}} = \frac{ \frac{8!}{7! 1!} \times \frac{5!}{3! 2!} }{ \frac{13!}{3! 10!} } = \frac{ 8 \times 10 }{ \frac{13 \times 12 \times 11}{6} } = \frac{40}{143}$

$\\$

Question 9: In a single throw of three dice, find the probability of getting the same number on all the three dice.

Three dices are thrown once.

Total number of possible outcomes is $\displaystyle 6^3 = 216$

$\displaystyle \text{Therefore, } n (S) = 216$

Let E be the event of getting same number on all the three dice

$\displaystyle \text{Therefore } E = 6 \text{ i.e.} \{(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5) (6,6,6) \}$

$\displaystyle n (E) =6$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6 }{ 216} = \frac{1}{36}$

$\\$

Question 10: Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.

Two dices are thrown.

Total number of possible outcomes is $\displaystyle 6^2 = 36$

$\displaystyle \text{Therefore, } n (S) = 36$

Let E be the event of getting sum on dice greater than 10

$\displaystyle \text{Therefore } E = 3 \text{ i.e.} \{ (5, 6), (6, 5), (6, 6) \}$

$\displaystyle n (E) =3$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{3 }{ 36} = \frac{1}{12}$

$\\$

Question 11: A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn
15:

(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 a heart nor a king

(vi) spade or an ace     (vii) neither an ace nor a king

(viii) neither a red card nor a queen       (ix) other than an ace

(x) a ten      (xi) a spade      (xii) a black card      (xiii) the seven of clubs

(xiv) jack      (xv) the ace of spades      (xvi) a queen      (xvii) a heart

(xviii) a red card

A card is drawn at random from a pack of 52 cards.

Total number of possible outcomes is $\displaystyle = 52$

$\displaystyle \text{Therefore, } n (S) = 52$

(i) a black king

Let E be the event of of getting a black king.

$\displaystyle \text{Therefore } E = 2 \text{ i.e.} \{ \text{King of Spade, King of Clubs} \}$

$\displaystyle n (E) =2$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{2 }{ 52} = \frac{1}{26}$

(ii) either a black card or a king

Let E be the event of of getting either a black card or a king

$\displaystyle \text{Therefore } E = 28 \text{ i.e.} \{ \text{13 spades, 13 clubs, king of hearts and diamonds} \}$

$\displaystyle n (E) =28$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{28 }{ 52} = \frac{7}{13}$

(iii) black and a king

Let E be the event of of getting either a black and a king

$\displaystyle \text{Therefore } E = 2 \text{ i.e.} \{ \text{king of spade and king of clubs} \}$

$\displaystyle n (E) =2$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{2 }{ 52} = \frac{1}{26}$

(iv) a jack, queen or a king

Let E be the event of of getting a jack, queen or a king

$\displaystyle \text{Therefore } E = 12 \text{ i.e.} \{ \text{4 jacks, 4 queens and 4 kings} \}$

$\displaystyle n (E) =12$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{12 }{ 52} = \frac{3}{13}$

(v) neither a heart nor a king

Let E be the event of of getting neither a heart nor a king

$\displaystyle \text{Therefore } E = 52 - 13 - 3 = 36 \{ \text{since we have 13 hearts, 3 kings each of spades, clubs and diamonds} \}$

$\displaystyle n (E) =36$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{36 }{ 52} = \frac{9}{13}$

(vi) spade or an ace

Let E be the event of of getting spade or an ace

$\displaystyle \text{Therefore } E = 16 \{ \text{13 spades and 3 aces each of hearts, diamonds and clubs} \}$

$\displaystyle n (E) =16$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{16 }{ 52} = \frac{4}{13}$

(vii) neither an ace nor a king

Let E be the event of of getting neither an ace nor a king

$\displaystyle \text{Therefore } E = 52-4-4 = 44 \{ \text{Since we have 4 aces and 4 kings} \}$

$\displaystyle n (E) =44$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{44 }{ 52} = \frac{11}{13}$

(viii) neither a red card nor a queen

Let E be the event of of getting neither a red card nor a queen

$\displaystyle \text{Therefore } E = 52 - 26 - 2 = 24 \{ \text{ Since we have 26 red cards of hearts and } \\ \text{ diamonds and 2 queens each of heart and diamond } \}$

$\displaystyle n (E) =24$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{24 }{ 52} = \frac{6}{13}$

(ix) other than an ace

Let E be the event of of getting other than an ace

$\displaystyle \text{Therefore } E = 52-4 = 48 \{ \text{Since we have 4 ace cards} \}$

$\displaystyle n (E) =48$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{48 }{ 52} = \frac{12}{13}$

(x) a ten

Let E be the event of of getting a ten

$\displaystyle \text{Therefore } E = 4 \{ \text{10 of spades, clubs, diamonds and hearts } \}$

$\displaystyle n (E) =4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 52} = \frac{1}{13}$

Let E be the event of of getting a spade

$\displaystyle \text{Therefore } E = 13 \{ \text{13 Spade cards} \}$

$\displaystyle n (E) =13$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{13 }{ 52} = \frac{1}{4}$

(xii) a black card

Let E be the event of of getting a black card

$\displaystyle \text{Therefore } E = 26 \{ \text{13 Spade cards and 13 Club cards} \}$

$\displaystyle n (E) =26$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{26 }{ 52} = \frac{1}{2}$

(xiii) the seven of clubs

Let E be the event of of getting the seven of clubs

$\displaystyle \text{Therefore } E = 1 \{ \text{7 of Clubs} \}$

$\displaystyle n (E) =1$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1 }{ 52}$

(xiv) jack

Let E be the event of of getting jack

$\displaystyle \text{Therefore } E = 4 \{ \text{4 Jacks} \}$

$\displaystyle n (E) =4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 52} = \frac{1}{13}$

(xv) the ace of spades

Let E be the event of of getting the ace of spades

$\displaystyle \text{Therefore } E = 1 \{ \text{Ace of Spade} \}$

$\displaystyle n (E) =1$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1 }{ 52}$

(xvi) a queen

Let E be the event of of getting a queen

$\displaystyle \text{Therefore } E = 4 \{ \text{4 Queens} \}$

$\displaystyle n (E) =4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 52} = \frac{1}{13}$

(xvii) a heart

Let E be the event of of getting a heart

$\displaystyle \text{Therefore } E = 13 \{ \text{13 hearts } \}$

$\displaystyle n (E) =13$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{13 }{ 52} = \frac{1}{4}$

(xviii) a red card

Let E be the event of of getting a red card

$\displaystyle \text{Therefore } E = 26 \{ \text{13 Diamonds and 13 Hearts} \}$

$\displaystyle n (E) =26$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{26 }{ 52} = \frac{1}{2}$

$\\$

Question 12: In shuffling a pack of 52 playing cards, four are accidently dropped; find the chance that the missing cards should be one from each suit.

We know that, from well shuffled pack of cards, 4 cards missed out total possible outcomes are

$\displaystyle \text{Therefore, } n (S) = ^{52} C_{4} = 270725$

Let E be the event that four missing cards are from each suite

$\displaystyle n(E) = ^{13} C_{1} \times ^{13} C_{1} \times ^{13} C_{1} \times ^{13} C_{1} = 13^4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{13^4 }{270725} = \frac{2197}{20825}$

$\\$

Question 13: From deck of 52 cards, four cards are drawn simultaneously, find the chance that they will be the four honors of the same suit.

$\displaystyle n (S) = ^{52} C_{4} = 270725$

Let E be the event that all the cards drawn are face cards of same suit.

$\displaystyle n(E) = 4 \times ^{4} C_{4} = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{270725}$

$\\$

Question 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?

$\displaystyle n (S) = 20$

Let E be Number of events of getting a multiples of 3 or 7 on the drawn ticket.

$\displaystyle n(E) = 8 \text{ i.e. } \{ 3,6,9,12,15,18,7,14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

$\\$

Question 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.

If three balls are drawn at random,

$\displaystyle n (S) = ^{18}C_{3} = 3 \times 17 \times 16$

Let E be the event that one is red, one is white and one is blue

$\displaystyle n(E) = ^{6}C_{1} \times ^{4}C_{1} \times ^{8}C_{1} = 6 \times 4 \times 8$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6 \times 4 \times 8 }{3 \times 17 \times 16} = \frac{4}{17}$

$\\$

Question 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 color

A bag contains 7 white,5 black and 4 red balls. If two balls are drawn at random

Total number of possible outcomes is $\displaystyle = ^{16}C_{2} = 120$

$\displaystyle \text{Therefore, } n (S) = 120$

(i) both the balls are white

Let E be Number of events of getting a multiples of 3 or 7 on the drawn ticket.

$\displaystyle n(E) = 8 \text{ i.e. } \{ 3,6,9,12,15,18,7,14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

(ii) one ball is black and the other red

Let E be Number of events of getting a multiples of 3 or 7 on the drawn ticket.

$\displaystyle n(E) = 8 \text{ i.e. } \{ 3,6,9,12,15,18,7,14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

(iii) both the balls are of the same color

Let E be Number of events of getting a multiples of 3 or 7 on the drawn ticket.

$\displaystyle n(E) = 8 \text{ i.e. } \{ 3,6,9,12,15,18,7,14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

$\\$

Question 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.

A bag contains 6 red,4 white and 8 blue balls. If three balls are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{18}C_{3} = 816$

$\displaystyle \text{Therefore, } n (S) = 816$

(i) one is red and two are white

Let E be Number of events of one is red and two are white.

$\displaystyle n(E) = ^{6}C_{1} \times ^{4}C_{1} = 36$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{36 }{816} = \frac{3}{68}$

(ii) two are blue and one is red

Let E be Number of events of two are blue and one is red.

$\displaystyle n(E) = ^{8}C_{2} \times ^{6}C_{1} = 168$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{168 }{816} = \frac{7}{34}$

(iii) one is red

Let E be Number of events of one is red.

$\displaystyle n(E) = ^{6}C_{1} \times ^{4}C_{1} \times ^{8}C_{1} +^{6}C_{1} \times ^{4}C_{2} + ^{6}C_{1} \times ^{8}C_{2}= 396$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{396 }{816} = \frac{33}{68}$

$\\$

Question 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?

Five cards are drawn from a pack of 52 cards.

Total number of possible outcomes is $\displaystyle = ^{52}C_{5} = 2598960$

$\displaystyle \text{Therefore, } n (S) = 2598960$

(i) just one ace

Let E be Number of events of just one ace.

$\displaystyle n(E) = ^{4}C_{1} \times ^{48}C_{4} = 778320$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{778320 }{2598960} = \frac{3243}{10829}$

(ii) at least one ace

Let E be Number of events of at least one ace.

$\displaystyle n(E) = ^{4}C_{1} \times ^{48}C_{4} + ^{4}C_{2} \times ^{48}C_{3} + ^{4}C_{3} \times ^{48}C_{4} + ^{4}C_{2} \times ^{48}C_{1} = 886656$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{886656 }{2598960} = \frac{18472}{54145}$

$\\$

Question 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?

The face cards are removed from a full pack. Out of the remaining 40 cards,4 are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{40}C_{4} = 91390$

$\displaystyle \text{Therefore, } n (S) = 91390$

Let E be the event that they belong to different suits

$\displaystyle n(E) = ^{10}C_{1} \times ^{10}C_{1} \times ^{10}C_{1} \times ^{10}C_{1} = 10000$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{10000}{91390} = \frac{1000}{9139}$

$\\$

Question 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 that it is a women.

There are four men and six women on the city councils. If one council member is selected for a committee at random.

Total number of possible outcomes is $\displaystyle = ^{10}C_{1} = 10$

$\displaystyle \text{Therefore, } n (S) = 10$

Let E be the event that it is a women.

$\displaystyle n(E) = ^{6}C_{1} =6$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6}{10} = \frac{3}{5}$

$\\$

Question 21: A box contains 100 bulbs, 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

A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection.

Total number of possible outcomes is $\displaystyle = ^{100}C_{10}$

$\displaystyle \text{Therefore, } n (S) = ^{100}C_{10}$

(i) all 10 are defective

Let E be the event when all 10 are defective.

$\displaystyle n(E) = ^{20}C_{10}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{20}C_{10}}{^{100}C_{10}}$

(ii) all 10 are good

Let E be the event when all 10 are good.

$\displaystyle n(E) = ^{80}C_{10}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{80}C_{10}}{^{100}C_{10}}$

(iii) at least one is defective

Let E be the event when at least one is defective.

Let E’ be the event when none of the bulds is defective.

$\displaystyle n(E') = ^{80}C_{10}$

$\displaystyle P (E') = \frac{n (E) }{ n (S)} = \frac{^{80}C_{10}}{^{100}C_{10}}$

$\text{Therefore } \displaystyle P (E) = 1 - P (E') = 1 - \frac{^{80}C_{10}}{^{100}C_{10}}$

(iv) none is defective

Let E be the event when all none is defective.

$\displaystyle n(E) = ^{80}C_{10}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{80}C_{10}}{^{100}C_{10}}$

$\\$

Question 22: Find the probability that in a random arrangement of the letters of the word ‘SOCIAL’ vowels come together.

Number of letters in ‘SOCIAL’ = 6

Therefore total possible outcomes of arranging the alphabets are $\displaystyle 6!$

$\displaystyle n (S) = 6!$

Let E be the event that vowels come together

Number of vowels in SOCIAL is A, I, O

So, number of ways to arrange them where, (A, I, O) come together

$\displaystyle n(E) = 4! \times 3!$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4! \times 3!}{6!} = \frac{1}{5}$

$\\$

Question 23:  The letters of the word ‘CLIFTON’ are placed at random in a row. What is the chance that two vowels come together?

Number of letters in ‘CLIFTON’ = 7

Therefore total possible outcomes of arranging the alphabets are $\displaystyle 7!$

$\displaystyle n (S) = 7!$

Let E be the event that vowels come together

Number of vowels in CLIFTON is I, O

So, number of ways to arrange them where, (I, O) come together

$\displaystyle n(E) = 6! \times 2!$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6! \times 2!}{7!} = \frac{2}{7}$

$\\$

Question 24:  The letters of the word ‘FORTUNATES’ are arranged at random in a row. What is the chance that the two ‘T’ come together.

Number of letters in ‘FORTUNATES’ = 10

Therefore total possible outcomes of arranging the alphabets are $\displaystyle 10!$

$\displaystyle n (S) = 10!$

Let E be the event that T’s come together

So, number of ways to arrange them where, (T, T) come together

$\displaystyle n(E) = 9! \times 2!$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{9! \times 2!}{10!} = \frac{1}{5}$

$\\$

Question 25:  A committee of two persons is selected form two men and two women; What is the probability that the committee will have (i) no man ? (ii) one man? (iii) two men?

Total number of persons $\displaystyle = 2 + 2 = 4$

Out of the 4 persons, two can be selected in $\displaystyle ^{4}C_{2}$ ways.

$\displaystyle n(S) = ^{4}C_{2}$

(i) no man

Let E be the event that committee will have no man

$\displaystyle n(E) = ^{2}C_{2}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{2}C_{2}}{^{4}C_{2}} = \frac{1}{\frac{4 \times 3}{2 \times 1}} = \frac{1}{6}$

(ii) one man

Let E be the event that committee will have one man

$\displaystyle n(E) = ^{2}C_{1} \times ^{2}C_{1}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{2}C_{1} \times ^{2}C_{1}}{^{4}C_{2}} = \frac{2 \times 2}{\frac{4 \times 3}{2 \times 1}} = \frac{1}{3}$

(iii) two men

Let E be the event that committee will have two man

$\displaystyle n(E) = ^{2}C_{2}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{2}C_{2}}{^{4}C_{2}} = \frac{1}{\frac{4 \times 3}{2 \times 1}} = \frac{1}{6}$

$\\$

Question 26:  If odds in favor of an event be 2:3, find the probability of occurrence of this event.

Odds in favor of an event be $\displaystyle 2:3$

Total number of possible outcomes $\displaystyle = 2k + 3k = 5k$

$\displaystyle n(S) = 5k$

Let E be the event that it occurs

$\displaystyle n(E) = 2k$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{2k}{5k} = \frac{2}{5}$

$\\$

Question 27: If odds against an event be 7:9, find the probability of non-occurrence of this event.

Odds against an event be $\displaystyle7:9$

Total number of possible outcomes $\displaystyle = 7k + 9k = 16k$

$\displaystyle n(S) = 16k$

Let E be the event that it does not occurs

$\displaystyle n(E) = 7k$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{7k}{16k} = \frac{7}{16}$

$\\$

Question 28: Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement. Find the probability that both the balls are of different colors.

Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement.

Total number of possible outcomes is $\displaystyle = ^{40}C_{4} = 91390$

$\displaystyle \text{Therefore, } n (S) = 91390$

Let E be the event that they belong to different suits

$\displaystyle n(E) = ^{10}C_{1} \times ^{10}C_{1} \times ^{10}C_{1} \times ^{10}C_{1} = 10000$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{10000}{91390} = \frac{1000}{9139}$

$\\$

Question 29: Two unbiased dice are thrown. Find the probability that:

(i) neither a doublet nor a total of 8 will appear

(ii) the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3

Two unbiased dice are thrown.

Total number of possible outcomes is $\displaystyle = ^{6}C_{1} \times ^{6}C_{1} = 36$

$\displaystyle \text{Therefore, } n (S) = 36$

(i) neither a doublet nor a total of 8 will appear

Let E’ be the event that a doublet or a total of 8 occurs

$\displaystyle E'= 10 \text{ i.e. } \{ (1,1), (2,2), (3,3), (4,4) ,(5,5), (6,6), (2,6), (6,2) ,(3,5), (5,3) \}$

$\displaystyle n(E') = 10$

$\displaystyle P (E') = \frac{n (E') }{ n (S)} = \frac{10}{36} = \frac{5}{18}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{5}{18} = \frac{13}{18}$

(ii) the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3

Let E’ be the event that sum of number obtain on the dice is either a multiple of 2 or 3, that is total should be 2,3,4,6,8,9,10,12

$\displaystyle E'= 24 \text{ i.e. } \{ (1,1) ,(1,2), (2,1), (1,3), (2,2), (3,1), \\ (1,5), (2,4), (3,3), (4,2), (5,1) ,(2,6), (3,5), (4,4) ,(5,3), (6,2), (3,6), \\ (4,5), (5,4), (6,3), (4,6), (5,5), (6,4) ,(6,6) \}$

$\displaystyle n(E') = 24$

$\displaystyle P (E') = \frac{n (E') }{ n (S)} = \frac{24}{36} = \frac{2}{3}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{2}{3} = \frac{1}{3}$

$\\$

Question 30: A bag contains 8 red, 3 white and 9 blue balls. If three balls are drawn at random, determine the probability that (i) all the three balls are blue balls (ii) all the balls are of different colors.

A bag contains 8 red, 3 white and 9 blue balls and three balls are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{20}C_{3} = 1140$

$\displaystyle \text{Therefore, } n (S) = 1140$

(i) all the three balls are blue balls

Let E be the event that all the three balls are blue balls

$\displaystyle n(E) = ^{9}C_{3} = 84$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{84}{1140} = \frac{7}{95}$

(ii) all the balls are of different colors

Let E be the event that all the balls are of different colors

$\displaystyle n(E) = ^{8}C_{1} \times ^{3}C_{1} \times ^{9}C_{1} =8 \times 3 \times 9 = 216$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{216}{1140} = \frac{18}{95}$

$\\$

Question 31: A bag contains 5 red, 5 white and 7 black balls. Two balls are drawn at random. What is the probability that both balls are red or both are black?

A bag contains 5 red, 5 white and 7 black balls. Two balls are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{18}C_{2} = 153$

$\displaystyle \text{Therefore, } n (S) = 153$

Let E be the event that both balls are red or both are black

$\displaystyle n(E) = ^{5}C_{2} + ^{7}C_{2} = 31$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{31}{153}$

$\\$

Question 32: If a letter is chosen at random from the English alphabet, find the probability that the letter is (i) a vowel (ii) a consonant

A letter is chosen at random from the English alphabet.

Total number of possible outcomes is $\displaystyle = ^{26}C_{1} = 26$

$\displaystyle \text{Therefore, } n (S) = 26$

(i) a vowel

Let E be the event that the letter is a vowel

Favorable outcomes are $\displaystyle a, e, i, o, u$

Total number of favorable outcomes = 5

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{5}{26}$

(ii) a consonant

Let E be the event that the letter is a consonant ( not a vowel)

Total number of favorable outcomes $\displaystyle = 26-5 = 21$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{21}{26}$

$\\$

Question 33: In a lottery, a person chooses six different numbers at random from 1 to 20, and if these six numbers match with six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?

A person chooses six different numbers at random from 1 to 20, and if these six numbers match with six numbers already fixed by the lottery committee, he wins the prize.

Total number of possible outcomes is $\displaystyle = ^{20}C_{6} = 38760$

$\displaystyle \text{Therefore, } n (S) = 38760$

Let E be the event that all six numbers match with the given number (as winning number is fixed).

$\displaystyle n(E) = 1$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1}{38760}$

$\\$

Question 34: 20 cards are numbered from 1 to 20. One card is drawn at random. What is the probability that the number on the cards is: (i) a multiple of 4? (ii) not a multiple of 4? (iii) odd? (iv) greater than 12? (v) divisible by 5? (vi) not a multiple of 6?

20 cards are numbered from 1 to 20. One card is drawn at random.

Total number of possible outcomes is $\displaystyle = ^{20}C_{1} = 20$

$\displaystyle \text{Therefore, } n (S) = 20$

(i) a multiple of 4

Let E be the event that the number on the drawn card is a multiple of 4

$\displaystyle n(E) = 5 \text{ i.e. } \{ 4, 8, 12, 16, 20 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{5 }{20} = \frac{1}{4}$

(ii) not a multiple of 4

Let E the event that the number on the drawn card is not a multiple of 4

$\displaystyle n(E') = 5 \text{ i.e. } \{ 4, 8, 12, 16, 20 \}$

$\displaystyle P (E') = \frac{n (E') }{ n (S)} = \frac{5 }{20} = \frac{1}{4}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{1}{4} = \frac{3}{4}$

(iii) odd

Let E be the event that the number on the drawn card is odd

$\displaystyle n(E) = 10 \text{ i.e. } \{ 1,3,5,7,9,11,13,15,17,19 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{10 }{20} = \frac{1}{2}$

(iv) greater than 12

Let E be the event that the number on the drawn card is greater than 12

$\displaystyle n(E) = 8 \text{ i.e. } \{ 13, 14, 15, 16, 17, 18, 19, 20 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

(v) divisible by 5

Let E be the event that the number on the drawn card is divisible by 5

$\displaystyle n(E) = 4 \text{ i.e. } \{ 5, 10, 15, 20 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{20} = \frac{1}{5}$

(vi) not a multiple of 6

Let E be the event that the number on the drawn card is divisible by 6

$\displaystyle n(E') = 3 \text{ i.e. } \{ 6, 12, 18 \}$

$\displaystyle P (E') = \frac{n (E) }{ n (S)} = \frac{3 }{20}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{3 }{20} = \frac{17 }{20}$

$\\$

Question 35: Two dice are thrown. Find the odds in favor of getting the sum (i) 4 (ii) 5 (iii) what are the odds against getting the sum 6?

Total number of possible outcomes is $\displaystyle = 6 \times 6 = 36$

$\displaystyle \text{Therefore, } n (S) = 36$

(i) sum is 4

Let E be the event that the sum is 4

$\displaystyle n(E) = 3 \text{ i.e. } \{ (1,3), (2, 2), (3, 1) \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{3 }{36} = \frac{1}{11}$

(ii) sum is 5

Let E be the event that the sum is 5

$\displaystyle n(E) = 4 \text{ i.e. } \{ (1,4), (2, 3), (3, 2), (4, 1) \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{36} = \frac{1}{9}$

(iii) what are the odds against getting the sum 6?

Let E’ be the event that the sum is 6

$\displaystyle n(E') = 5 \text{ i.e. } \{ (1,5), (2, 4), (3, 3), (4, 2), (5, 1) \}$

$\displaystyle P (E') = \frac{n (E) }{ n (S)} = \frac{5 }{36}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{5 }{36} = \frac{31 }{36}$

$\\$

Question 36: What are the odds in favor of getting a spade if a card is drawn from a well-shuffled deck of cards? What are the odds in favor of getting a king?

A card is drawn from a well-shuffled deck of cards.

Total number of possible outcomes is $\displaystyle = ^{52}C_{1} = 52$

$\displaystyle \text{Therefore, } n (S) = 52$

(i) Let E be the event that we get a spade

$\displaystyle n(E) = 13 \text{ i.e. } \{ 13 \text{ Spade cards } \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{13 }{52} = \frac{1}{4}$

Therefore, probability of event E’ is

$\displaystyle P(E') = 1 - P(E) = 1 - \frac{1 }{4} = \frac{3 }{4}$

$\displaystyle \text{Odds in favour of getting a spade is } \frac{P(E)}{P(E')} = \frac{\frac{1}{4}}{\frac{3 }{4}} =1:3$

(ii) Let E be the event that we get a king

$\displaystyle n(E) = 4 \text{ i.e. } \{ 4 \text{ Kings } \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{52} = \frac{1}{13}$

Therefore, probability of event E’ is

$\displaystyle P(E') = 1 - P(E) = 1 - \frac{1 }{13} = \frac{12 }{13}$

$\displaystyle \text{Odds in favour of getting a king is } \frac{P(E)}{P(E')} = \frac{\frac{1}{13}}{\frac{12 }{13}} =1:12$

$\\$

Question 37: A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random. From the box, what is the probability that (i) all are blue? (ii) at least one is green?

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{60}C_{5}$

$\displaystyle \text{Therefore, } n (S) = ^{60}C_{5}$

(i) all are blue?

Let E be the event that we get all blue

5 blue marbles out of 20 can be drawn in $\displaystyle ^{20}C_{5}$ ways

$\displaystyle n(E) = ^{20}C_{5}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{20}C_{5} }{^{60}C_{5}} = \frac{20 \times 19 \times 18 \times 17 \times 16}{60\times 59\times 58\times 57 \times 56} = \frac{ 2 \times 17 }{59 \times 29 \times 7} = \frac{34}{11977}$

(ii) at least one is green

Let E’ be the event that no green marble is chosen

5 non-green marbles out of 30 can be drawn in $\displaystyle ^{30}C_{5}$ ways

$\displaystyle n(E') = ^{30}C_{5}$

$\displaystyle P (E') = \frac{n (E) }{ n (S)} = \frac{^{30}C_{5} }{^{60}C_{5}}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{^{30}C_{5} }{^{60}C_{5}} = 1 - \frac{30 \times 39 \times 38 \times 37 \times 36 }{60 \times 59 \times 58 \times 57 \times 56 } \\ \\ \\ = 1 - \frac{9 \times 13}{4 \times 59 \times 19} = 1- \frac{117}{4484} = \frac{4367}{4484}$

$\\$

Question 38: A box contains 6 red marbles numbered 1 through 6 and 4 white marbles numbered from 12 through 15. Find the probability that a marble drawn is (i) white (ii) white and odd numbered (iii) even numbered (iv) red or even numbered.

A box contains 6 red marbles numbered 1 through 6 and 4 white marbles numbered from 12 through 15. One marble is drawn.

Total number of possible outcomes is $\displaystyle = ^{10}C_{1} = 10$

$\displaystyle \text{Therefore, } n (S) = 10$

(i) white

Let E be the event of getting white marble

$\displaystyle n(E) = ^{4}C_{1} = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{10} = \frac{2}{5}$

(ii) white and odd numbered

Let E be the event of getting white and odd numbered marble

$\displaystyle n(E) = 2 \text{ i.e. } \{ 13, 15 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{2 }{10} = \frac{1}{5}$

(iii) even numbered

Let E be the event of getting even numbered marble

$\displaystyle n(E) = 5 \text{ i.e. } \{ 2, 4, 6, 12, 14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{5 }{10} = \frac{1}{2}$

(iv) red or even numbered

Let E be the event of getting red or even numbered marble

$\displaystyle n(E) = 8 \text{ i.e. } \{1, 2, 3, 4, 5, 6, 12, 14\}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{10} = \frac{4}{5}$

$\\$

Question 39: A class consists of 10 boys and 8 girls. 3 students are selected at random. What is the probability that the selected group has (i) all boys? (ii) all girls? (iii) 1 boy and 2 girls? (iv) at least one girl? (v) at most one girl?

A class consists of 10 boys and 8 girls. 3 students are selected at random.

Total number of possible outcomes is $\displaystyle = ^{18}C_{3} = 816$

$\displaystyle \text{Therefore, } n (S) = 816$

(i) all boys

Let E be the event of getting all boys

$\displaystyle n(E) = ^{10}C_{3} =120$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{120 }{816} = \frac{5}{34}$

(ii) all girls

Let E be the event of getting all girls

$\displaystyle n(E) = ^{8}C_{3} =56$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{56 }{816} = \frac{7}{102}$

(iii) 1 boy and 2 girls

Let E be the event of getting 1 boy and 2 girls

$\displaystyle n(E) = ^{10}C_{1} \times ^{8}C_{2} =280$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{280 }{816} = \frac{35}{102}$

(iv) at least one girl

Let E be the event of getting at least one girl

$\displaystyle n(E) = ^{10}C_{2} \times ^{8}C_{1} + ^{10}C_{1} \times ^{8}C_{2} + ^{10}C_{0} \times ^{8}C_{3}=696$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{696 }{816} = \frac{29}{34}$

(v) at most one girl

Let E be the event of getting atmost one girl

$\displaystyle n(E) = ^{10}C_{2} \times ^{8}C_{1} + ^{10}C_{3} \times ^{8}C_{0} =480$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{480 }{816} = \frac{10}{17}$

$\\$

Question 40: Five cards are drawn from a well-shuffled pack of 52 cards. Find the probability that all the
five cards are hearts.

Five cards are drawn from a well-shuffled pack of 52 cards.

Total number of possible outcomes is $\displaystyle = ^{52}C_{5} =2598960$

$\displaystyle \text{Therefore, } n (S) = 2598960$

Let E be the event that all the five cards are hearts.

$\displaystyle n(E) = ^{13}C_{5} = 1287$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1287 }{2598960} = \frac{33}{66640}$

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Question 41: A bag contains tickets numbered from 1 to 20. Two tickets are drawn. Find the probability that (i) both the tickets have prime numbers on them (ii) on one there is a prime number and on the other there is a multiple of 4.

A bag contains tickets numbered from 1 to 20. Two tickets are drawn.

Total number of possible outcomes is $\displaystyle = ^{20}C_{2} =190$

$\displaystyle \text{Therefore, } n (S) = 190$

(i) both the tickets have prime numbers on them

Let E be the event that both the tickets have prime numbers on them.

$\displaystyle E= \{ 2,3,5,7,11,13,17,19 \}$

$\displaystyle n(E) = ^{8}C_{2} = 28$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{28 }{190} = \frac{28}{190}$

(ii) one there is a prime number and on the other there is a multiple of 4

Let E be the event that one there is a prime number and on the other there is a multiple of 4.

$\displaystyle E= \{ 2,3,5,7,11,13,17,19 \}$ for prime number

$\displaystyle E= \{ 4,8,12,16,20 \}$ for multiple of 4

$\displaystyle n(E) = ^{8}C_{1} \times ^{5}C_{1}= 40$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{40 }{190} = \frac{4}{19}$

$\\$

Question 42: An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that (i) both the balls are red (ii) one ball is red and the other is black (iii) one ball is white.

An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{18}C_{2} =105$

$\displaystyle \text{Therefore, } n (S) = 105$

(i) both the balls are red

Let E be the event that both the balls are red

$\displaystyle n(E) = ^{3}C_{2} = 3$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{3 }{105} = \frac{1}{35}$

(ii) one ball is red and the other is black

Let E be the event that one ball is red and the other is black

$\displaystyle n(E) = ^{3}C_{1} \times ^{5}C_{1} = 15$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{15}{105} = \frac{1}{7}$

(iii) one ball is white

Let E be the event that one ball is white

$\displaystyle n(E) = ^{8}C_{1} \times ^{7}C_{1} = 56$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{56}{105} = \frac{8}{15}$

$\\$

Question 43: A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number.

Two dices are thrown once.

Total number of possible outcomes is $\displaystyle 6^2 = 36$

$\displaystyle \text{Therefore, } n (S) = 36$

Let E be the event of B throwing a no. higher than 9

No. of favorable outcomes $\displaystyle = 6 \{ (4, 6), (5, 5), (6, 4), (5, 6) ,(6, 5), (6, 6) \}$

$\displaystyle n(E) = 6$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6}{36} = \frac{1}{6}$

$\\$

Question 44: In a hand at Whist, what is the probability that four kings are held by a specified player?

We know that 13 cards are delivered to a hand at whist.

Total number of possible outcomes is $\displaystyle ^{52}C_{13}$

$\displaystyle \text{Therefore, } n (S) = ^{52}C_{13}$

Since 4 kings are held by a specified player, 9 more cards are to be delivered to him from the remaining 48 cards

No of favorable ways $\displaystyle = ^{48}C_{9}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{48}C_{9}}{^{52}C_{13}} = \frac{13 \times 12 \times 11 \times 10}{52 \times 51 \times 50 \times 49 } = \frac{11}{4165}$

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Question 45: Find the probability that in a random arrangement of the letters of the word ‘UNIVERSITY’, the two I’s do not come together.

Let S be the sample space.

Then, $\displaystyle n(S) =$ Total number of ways in which the letters of the word  $\displaystyle \text{UNIVERSITY can be arranged} = \frac{10!}{2!}$

Let E be the event of arranging the letters of the word ‘UNIVERSITY’ such that 2 I’s remain together.

Number of ways where the two Is are always together $\displaystyle = 9!$

$\displaystyle \text{Therefore the number of ways where the 2Is are not together } = \frac{10!}{2!} - 9!$

$\displaystyle \text{Therefore the probability } = \frac{\frac{10!}{2!} - 9!}{\frac{10!}{2!}} = \frac{4}{5}$

$\\$