OP Malhotra Class 11 Maths Solutions Chapter 22 Probability Ex 22(f)

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S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(f)

Question 1.
(i) In a single throw of two coins, find the probability of getting both heads or both tails.
(ii) A dice is thrown twice. Find the probability that the sum of the two numbers obtain is 5 or 7 ?
(iii) Two dice are tossed once. Find the probability of getting an even number on first die or a total of 8.
Solution:
(i) In a single throw of two coins
Then sample space S = {HH, HT, TH, TT}
∴ Total no. of exhaustive cases = 4
P (both heads) = \(\frac { 1 }{ 4 }\) {HH}
P (both tails) = \(\frac { 1 }{ 4 }\) {TT}
Thus required probability = P (both heads) + P (tails) = \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

(ii) Here total no. of outcomes = 6² = 36 = n (S)
Let A : sum of two numbers be 5 = {(1, 4), (2, 3), (3, 2), (4, 1)} ∴ n (A) = 4
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{4}{36}\)
B : event that sum of numbers be 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} ∴ n (B) = 6
Thus P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{6}{36}\)
∴ required probability of getting either sum 5 or 7 = P (A or B) = P (A) + P (B) = \(\frac { 4 }{ 36 }\) + \(\frac { 6 }{ 36 }\) = \(\frac { 10 }{ 36 }\) = \(\frac { 5 }{ 18 }\)
[since A and B are mutually exclusive events as A ∩ B = Φ) P (A ∩ B) = 0]

(iii) Here, total no. of exhaustive cases = n (S) = 6² = 36
Let A : event of getting an even number on first die
= {(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(A) = 18

Question 2.
If the probability of a horse A winning a race is \(\frac { 1 }{ 5 }\) and the probability of horse B winning the same race is \(\frac { 1 }{ 4 }\), what is the probability that one of the horses will win ?
Solution:
Given P (horse A winning a race) = \(\frac { 1 }{ 5 }\); P (horse B winning a race) = \(\frac { 1 }{ 4 }\)
∴ required probability = P (A) + P(B) = \(\frac { 1 }{ 5 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 9 }{ 20 }\)

Question 3.
In a single throw of two dice, what is the probability of obtaining a total of 9 or 11 ?
Solution:
In a single throw of two dice
Then total no. of exhaustive cases = 6² = 36 = n (S)
Let A : event of obtaining a total of 9 = {(3, 6), (4, 5), (5,4), (6, 3)} ∴n (A) = 4
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 26 }\)
Let B : obtaining a total of 11 = {(5, 6), (6, 5)} ∴ n(B) = 2
Thus P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 36 }\)
Here A ∩ B = Φ ⇒ P(A ∩ B) = 0
Thus required probability of obtaining a total of 9 or 11 = P (A or B)
= P (A) + P (B) – P (A ∩ B) = \(\frac { 4 }{ 36 }\) + \(\frac { 2 }{ 36 }\) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Question 4.
In a group there are 2 men and 3 women. 3 persons are selected at random from the group. Find the probability that 1 man and 2 women or 2 men and 1 woman are selected.
Solution:
Total no. of persons = 2 + 3 = 5
∴ Total no. of ways of selecting 3 persons out of 5 = ⁵C₃
Let A : event that 1 man and 2 women are selected ∴ n(A) = ²C₁ × ³C₂

Question 5.
In a class of 25 students with roll numbers 1 to 25, a student is picked up at random to answer a question. Find the probability that the roll number of the selected student is either a multiple of 5 or 7 ?
Solution:
Total exhaustive cases = 25 = n (S)
Let A : event that roll no. of selected student be a multiple of 5 = {5, 10, 15, 20, 25}
B : event that roll no. of selected student be a multiple of 7 = {7, 14,21}
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{5}{25}\); P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{3}{25}\)
Thus required probability = P(A or B) = P(A) + P(B) = \(\frac { 5 }{ 25 }\) + \(\frac { 3 }{ 25 }\) = \(\frac { 8 }{ 25 }\)

Question 6.
If chance of A, winning a certain race be \(\frac { 1 }{ 6 }\) and the chance of B winning it is \(\frac { 1 }{ 3 }\), what is the chance that neither should win ?
Solution:
Given P(A’s Winning) = \(\frac { 1 }{ 6 }\), P(B’s Winning) = \(\frac { 1 }{ 3 }\)
∴ P (neither A nor B wins)

Question 7.
Discuss and criticise the following : P(A) = \(\frac { 2 }{ 3 }\), P(B) = \(\frac { 1 }{ 4 }\), P (C) = \(\frac { 1 }{ 3 }\) are the probabilities of three mutually exclusive events A, B and C.
Solution:
Since A, B and C are mutually exclusive events.
Now P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 3 }\) = \(\frac { 5 }{ 4 }\) > 1
But P(A ∪ B ∪ C) ≤ 1 Thus, given statement is false.

Question 8.
E and F are two events associated with a random experiment for which P (F) = 0.35, P (E or F) = 0.85, P (E and F) = 0.15. Find P (E).
Solution;
Given P (F) = 0.35 ; P (E or F) = 0.85 and P (E and F) = 0.15
We know that P (E ∪ F) = P (E) + P (F) – P (E ∩ F)
⇒ 0.85 = P(E) + 0.35 – 0.15
⇒ P (E) = 0.85 – 0.20 = 0.65

Question 9.
(i) Two events A and B have probabilities 0.25 and 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Find the probability that neither A nor B occurs.
(ii) The probability of an event A occurring is 0.5 and of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B occurring.
Solution:
(i) Given P (A) = 0.25 ; P (B) = 0.50 and P (A ∩ B) = 0.14
P (neither A nor B) = P(\(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\)) = P \(P(\overline{A \cup B})\) = 1 – P(A ∪ B)
= 1 – [P (A) + P (B) – P (A ∩ B)] = 1 – [0.25 + 0.50 – 0.14] = 1 – 0.61 = 0.39

(ii) Given P (A) = 0.5 ; P (B) = 0.3
Since A and B are mutually exclusive events. ∴ A ∩ B = Φ ⇒ P (A ∩ B) = 0
P (neither A nor B) = P(\(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\)) = P \(P(\overline{A \cup B})\) = 1 – P(A ∪ B)
= 1 – [P (A) + P (B) – P (A ∩ B)] = 1 – 0.5 – 0.3 – 0 = 0.2

Question 10.
(i) A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product of the numbers is event ?
(ii) A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
Solution:
(i) Total no. of ways of drawing 2 tickets from 25 tickets = ²⁵C₂
The product of numbers is even when either both are even or one is even and the other is odd
Let A : event that both are even and there are 12 even numbers from 1 to 25
∴ n (A) = Total no. of ways of drawing 2 tickets out of 12 = ¹²C₂
Let B : event that one is even and other is odd since there are 12 even and 13 odd numbers from 1 to 25.

(ii) Total no. of balls = 7 + 5 + 4 = 16
n (S) = Total no. of ways of drawing four balls out of 16 without replacement = ¹⁶C₄
Drawing atleast three black balls means drawing 3 black balls out of 5 and one ball from remaining 11 balls or 4 black balls out of 5.
Thus required probability = \(\frac{{ }^5 C_3 \times{ }^{11} C_1+{ }^5 C_4}{{ }^{16} C_4}\) = \(\frac{\frac{5 \times 4}{2} \times 11+5}{\frac{16 \times 15 \times 14 \times 13}{24}}\) = \(\frac { 23 }{ 364 }\)

Question 11.
(i) A and B are two mutually exclusive events of an experiment.
If P(not A) = 0.75, P(A ∪ B) = 0.65 and P(B) = p, find the value of p.
(ii) A and B are two mutually exclusive events. If P (A) = 0.5 and P \((\overline{\mathbf{B}})\) = 0.6, find P (A or B).
Solution:
(i) Given A and B are mutually exclusive events.
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
Given P (not A) = 0.65 ⇒ 1 – P (A) = 0.65 ⇒ P (A) = 0.35
P(A ∪ B) = 0.65 and P (B) =p
We know that P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ 0.65 = 0.35 + p – 0 ⇒ p = 0.30

(ii) Given A and B are two mutually exclusive events.
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
Given P (A) = 0.5 ; P\((\overline{\mathbf{B}})\) = 0.6 ⇒ 1 – P (B) = 0.6 ⇒ P (B) = 0.4
∴ P (A or B) = P (A) + P (B) – P (A ∩ B) = 0.5 + 0.4 – 0 = 0.9

Question 12.
(i) A, B and C are three mutually exclusive and exhaustive events associated with a random experiment. Find P (A) given that P (B) = \(\frac { 3 }{ 2 }\) P (A) and P (C) = \(\frac { 1 }{ 2 }\) P (B).
Solution:
(i) Given A, B and C are mutually exclusive, exhaustive events
∴ P(A) + P(B) + P(C) = 1 …(1)
Given P(B) = \(\frac { 3 }{ 2 }\) P(A) and P (C) =\(\frac { 1 }{ 2 }\) P(B) = \(\frac { 1 }{ 2 }\) × \(\frac { 3 }{ 2 }\)P(A) = \(\frac { 3 }{ 4 }\) P (A)
∴ from (1); P(A) + \(\frac { 3 }{ 2 }\)P (A) + \(\frac { 3 }{ 4 }\) P(A) = 1
\(\left[1+\frac{3}{2}+\frac{3}{4}\right] \mathrm{P}(\mathrm{A})=1\)
\(\Rightarrow\left(\frac{4+6+3}{4}\right) P(A)=1\)
⇒ P (A) = \(\frac { 4 }{ 13 }\)

(ii) Since A, B and C are mutually exclusive and exhaustive events.
∴ P (A) + P (B) + P (C) = 1
Given P (A) = 2 P (B) = 3 P (C) …(1)

Question 13.
(i) In a single throw of two dice, find the probability that neither a doublet nor a total of a 10 will appear.
(ii) Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 3 nor a multiple of 4.
(iii) Two dice are thrown together ; what is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 5.
Solution:
(i) In a single throw of two dice, total exhaustive cases = 6² = 36
Let A: getting a doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} ∴ n (A) = 6
B : getting a total of 10 = {(4, 6), (5, 5), (6, 4)} ∴n (B) = 3
Thus P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 6 }{ 36 }\); P (B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 3 }{ 36 }\)
and A ∩ B = {(5, 5)} ∴ n (A ∩ B) = 1
Thus P (A ∩ B) = \(\frac{n(\mathrm{~A} \cup \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 36 }\)
∴ P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac { 6 }{ 36 }\) + \(\frac { 3 }{ 36 }\) – \(\frac { 1 }{ 36 }\) = \(\frac { 8 }{ 36 }\) = \(\frac { 2 }{ 9 }\)
Thus, required probability = P (neither A nor B) = \(\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})\) = \(P(\overline{A \cup B})\)

(ii) Total exhaustive cases = n (S) = 6² = 36
Let A : event that the sum of the numbers obtained on two dice is multiple of 3
= {(U 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} ∴ n (A)= 12
Let B = event that sum of the numbers on two dice is multiple of 4
= {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)} ∴ n (B) = 9
∴ A ∩ B = {(6, 6)} ⇒ n (A ∩ B) = 1
Thus, P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{12}{36}\); P (B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{9}{36}\) and P (A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 36 }\)
∴ P(A ∪ B) = P (A) + P (B) = P (A ∩ B) = \(\frac { 12 }{ 36 }\) + \(\frac { 9 }{ 36 }\) – \(\frac { 1 }{ 36 }\) = \(\frac { 20 }{ 36 }\) = \(\frac { 5 }{ 9 }\)
∴ required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A u B) = 1 – \(\frac { 5 }{ 9 }\) = \(\frac { 4 }{ 9 }\)

(iii) When two dice are thrown together
Then total no. of exhaustive cases = n (S) = 6² = 36
Let A : event that the sum of the numbers on the two faces is divisible by 3
= {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} ∴ n (A) = 12
Let B : event that the sum of numbers on two faces is divisible by 5
= {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)} ∴ n (B) = 7
and A ∩ B = Φ ⇒ P (A ∩ B) = 0
Thus, P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – 0 = \(\frac { 12 }{ 36 }\) + \(\frac { 7 }{ 36 }\) = \(\frac { 19 }{ 36 }\)
∴ Required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A ∪ B) = 1 – \(\frac { 19 }{ 36 }\) = \(\frac { 17 }{ 36 }\)

Question 14.
In a given race, the odds in favour of horses A, B, C and D are 1 : 3, 1 : 4, 1 : 5 and 1 : 6 respectively. Find the probability that one of them wins the race.
Solution:
Given odds in favour of horse A be 1 : 3
∴ P (horse A’s winning) = \(\frac{1}{1+3}\) = \(\frac{1}{4}\)
odds in favour of horse B be 1 : 4
∴ P (horse B’s winning) = \(\frac{1}{1+4}\) = \(\frac{1}{5}\)
odds in favour of horse C be 1 : 5
∴ P (horse C’s winning) = \(\frac{1}{1+5}\) = \(\frac{1}{6}\)
and odds in favour of horse D be 1 : 6
∴ D (horse D’s winning) = \(\frac{1}{1+6}\) = \(\frac{1}{7}\)
∴ Required probability that one of them will win the race = P (A) + P (B) + P (C) + P (D)
[since A, B, C and D are mutually exclusive and exhaustive events]
= \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{1}{6}\) + \(\frac{1}{7}\) = \(\frac{105+84+70+60}{420}\) = \(\frac{319}{420}\)

Question 15.
100 students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has failed in both examinations.
Solution:
Let A : event that students pass the fist examination
B : event that students pass the second exam.
Then P(A) = \(\frac{60}{100}\); P (B) = \(\frac{5}{100}\); P(A ∩ B) = \(\frac{60}{100}\) + \(\frac{50}{100}\) – \(\frac{30}{100}\) = \(\frac{80}{100}\)
Thus required probability that selected student failed in both exams = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\)
= 1 – P (A ∪ B) = 1 – \(\frac{80}{100}\) = \(\frac{20}{100}\) = 0.2

Question 16.
(i) A card is drawn from a deck of 52 can is. Find the probability of getting an ace or a spade card.
(ii) From a well shuffled deck of 52 cards, 4 cards are drawn at random. What is the probability that all the drawn cards are of the same colour.
Solution:
(i) Total no. of exhaustive cases = 52
Let A : event of getting an ace card ∴ n (A) = 4
B : getting a spade card ∴ n (B) = 13
A ∩ B : getting an ace of spade
∴ n (A ∩ B) = 1
Thus P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\)
= \(\frac{4}{52}\) + \(\frac{13}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(ii) n(S) = Total no. of ways of drawing 4 cards out of 52 = ⁵²C₄
All drawing cards are of same colour means all cards are either red or black.
Let A : drawing four red cards.
∴ n (A) = Total no. of ways of drawing four cards out of 26 = ²⁶C₄
Let B : drawing four black cards
∴ n (B) = Total no. of ways of drawing four cards out of 26 = ²⁶C₄
Thus required probability =P (A) + P (B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{2 \times{ }^{26} \mathrm{C}_4}{{ }^{52} \mathrm{C}_4}\)
= 2 × \(\frac{26 \times 25 \times 24 \times 23}{52 \times 51 \times 50 \times 49}\) = \(\frac{92}{833}\)

Question 17.
(i) A card is drawn at random from a well shuffled pack of cards. What is the probability that it is a heart or a queen ?
(ii) A card is drawn at random from a well shuffled pack of 52 cards. Find the probability it is neither a king nor a heart ?
Solution:
(i) Given total exhaustive cases = 52 = n (S)
Let A : event that heart card is drawing ∴ n (A) = 13
and B : drawing a queen card ∴ n (B) = 4
A ∩ B : drawing a queen of heart ∴ n (A ∩ B) = 1
Thus required probability = P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
= \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac{13}{52}\) + \(\frac{4}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(ii) Total no. of exhaustive cards = n (S) = 52
Let A : drawing a king card ∴ n (A) = 4
and B : drawing a heart card ∴ n (B) = 13
A ∩ B : drawing a king card of heart ∴ n (A ∩ B) = 1
Thus P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac{4}{52}\) + \(\frac{13}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)
∴ Required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A ∪ B) = 1 – \(\frac{4}{13}\) = \(\frac{9}{13}\)

Question 18.
A card is drawn from a pack of 52 cars. Find the probability of getting a king or a heart or a red card.
Solution:
Total no. of outcomes = n (S) = 52
Let A : drawing a king card ∴ n (A) = 4
B : drawing a heart card ∴ n (B) = 13
C : drawing a red card ∴ n (C) = 26
A ∩ B : drawing a king of heart card
∴ n( A ∩ B) = 1
B ∩ C : drawing a heart card
∴ n( B ∩ C) = 13
A ∩ C : drawing a king of red card
∴ n( A ∩ C) = 2
∴ A ∩ B ∩ C : drawing a king of heart card
∴ n(A ∩ B ∩ C) = 1
Required probability = P(A ∪ B ∪ C) = P (A) + P (B) + P(C) – P (A ∩ B) -P (B ∩ C) – P (A ∩ C) + P(A ∩ B ∩ C)
= \(\frac{4}{52}\) + \(\frac{13}{52}\) + \(\frac{26}{52}\) –\(\frac{1}{52}\) – \(\frac{13}{52}\) – \(\frac{2}{52}\) + \(\frac{1}{52}\) = \(\frac{28}{52}\) = \(\frac{7}{13}\)

Question 19.
There are three events A, B, C one of which must and only one can happen. The odds are 8 to 3 against A, 5 to 2 against B ; find the odds against C.
Solution:
Given odds against A be 8 : 3.
∴ P (A) = \(\frac{3}{8+3}\) + \(\frac{3}{11}\)
Also, odds against B be 5 : 2
Since out of three events A, B and C, one of which must and only one can happen ∴ A, B and C are mutually exclusive and exhaustive events.
∴ P (A) + P (B) + P (C) = 1
⇒ P (C) = 1 – \(\frac{3}{11}\) – \(\frac{2}{7}\) = \(\frac{77-21-22}{77}\) = \(\frac{34}{77}\)
\(\mathrm{P}(\overline{\mathrm{C}})\) = 1 – P (C) = 1 – \(\frac{34}{77}\) = \(\frac{43}{77}\)
Thus required odds against C be \(\mathrm{P}(\overline{\mathrm{C}})\) : P (C) i.e. \(\frac{43}{77}\) : \(\frac{34}{77}\) i.e. 43 : 34.

Question 20.
In a group of students, there are 3 boys and 3 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Solution:
Given total no. of students = 3 + 3 = 6
∴ n (S) = Total no. of ways of selecting 4 students out of 6 = ⁶C₄
Let A : selecting 3 boys and 1 girl
∴ n (A) = Total no. of ways of selecting 3 boys out of 3 and 1 girl out of 3 = ³C₃ × ³C₁
Let B : selecting 1 boy and 3 girls
∴ n (B) = Total no. of ways of selecting 1 boy out of 3 and 3 girls out of 3 = ³C₁ × ³C₃
Thus required probability = P (A ∪ B) = P
(A) + P (B) = \(\frac{{ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_3}{{ }^6 \mathrm{C}_4}\) = \(\frac{1 \times 3+3 \times 1}{\frac{6 \times 5}{2}}\) = \(\frac{6}{15}\) =  \(\frac{2}{5}\)