Class 11: Probability – Exercise 33.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1:

(a) If A and B be mutually exclusive events associated with a random experiment such that $\displaystyle P(A) = 0.4 \text{ and } P(B) =0.5$ then find:

(i) $\displaystyle P(A \cup B)$ (ii) $\displaystyle P( \overline{A} \cap \overline{B})$ (iii) $\displaystyle P(\overline{A} \cap B)$ (iv) $\displaystyle P(A \cap \overline{B})$

(b) A and B are two events such that $\displaystyle P(A)=0.54, \ P (B)=0.69$ and $\displaystyle P (A \cap B) = 0.35$

Find (i) $\displaystyle P(A \cup B)$ (ii) $\displaystyle P(\overline{A} \cup \overline{B})$ (iii) $\displaystyle P( A \cap \overline{B})$ (iv) $\displaystyle P(B \cap A)$

	$\displaystyle P(A)$	$\displaystyle P(B)$	$\displaystyle P(A \cap B)$	$\displaystyle P(A \cup B)$
i)	$\displaystyle \frac{1}{3}$	$\displaystyle \frac{1}{5}$	$\displaystyle \frac{1}{15}$	$\displaystyle \cdots$
ii)	$\displaystyle 0.35$	$\displaystyle \cdots$	$\displaystyle 0.25$	$\displaystyle 0.6$
iii)	$\displaystyle 0.5$	$\displaystyle 0.35$	$\displaystyle \cdots$	$\displaystyle 0.7$

Answer:

(a)

Given $\displaystyle P(A) = 0.4 \text{ and } P(B) =0.5$

A and B be mutually exclusive events, therefore $\displaystyle P( \cap B) = 0$

$\displaystyle \text{(i) } P(A \cup B) = P(A)+P(B) = 0.4 +0.5 = 0.9$

$\displaystyle \text{(ii) } P( \overline{A} \cap \overline{B}) = 1 - P( A \cup B) = 1 - 0.9 = 0.1$

$\displaystyle \text{(iii) } P(\overline{A} \cap B) = P(B) - P(A \cap B) = 0.5-0=0.5$

$\displaystyle \text{(iv) } P(A \cap \overline{B}) = P(A) - P(A \cap B) 0.4-0=0.4$

(b)

Given $\displaystyle P(A)=0.54, \ P (B)=0.69 \text{ and } P (A \cap B) = 0.35$

$\displaystyle \text{(i) } P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.54+0.69-0.35 = 1.23-0.35 = 0.88$

$\displaystyle \text{(ii) } P(\overline{A} \cup \overline{B}) = 1 - P(A \cup B) = 1 - 0.88 = 0.12$

$\displaystyle \text{(iii) } P( A \cap \overline{B}) = P(A)- P(A \cap B) = 0.54-0.35 = 0.19$

$\displaystyle \text{(iv) } P(B \cap A) = P(B) - P(A \cap B) = 0.69-0.35=0.34$

(c)

$\displaystyle \text{(i) } P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15} = \frac{7}{15}$

$\displaystyle \text{(ii) } P(A \cup B) = P(A) + P(B) - P(A \cap B) \\ \\ \Rightarrow P(B) = P(A \cup B) - P(A) + P(A \cap B) = 0.6-0.35+0.25 = 0.5$

$\displaystyle \text{(iii) } P(A \cup B) = P(A) + P(B) - P(A \cap B) \\ \\ \Rightarrow P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5+0.35 - 0.7 = 0.15$

$\\$

Question 2: If A and B are two events associated with a random experiment such that $\displaystyle P(A) = 0.3, \ P(B) = 0.4 \text{ and } P (A \cup B) = 0.5 \text{ find } P (A \cap B)$

Answer:

$\displaystyle \text{Given } P(A) = 0.3, \ P(B) = 0.4 \text{ and } P (A \cup B) = 0.5 \text{ find } P (A \cap B)$

$\displaystyle P(A \cup B) = P(A)+P(B) - P(A \cap B)$

$\displaystyle \Rightarrow P(A \cap B) = P(A)+P(B) - P(A \cup B) = 0.3+0.4-0.5 = 0.2$

$\\$

Question 3: If A and B are two events associated with a random experiment such that $\displaystyle P(A) = 0.5, P(B) = 0.3$ and $\displaystyle P (A \cap B) =0.2,$ find $\displaystyle P (A \cup B).$

Answer:

$\displaystyle \text{Given } P(A) = 0.5, P(B) = 0.3 \text{ and } P (A \cap B) =0.2, \text{ find } P (A \cup B).$

$\displaystyle P(A \cup B) = P(A)+P(B) - P(A \cap B) = 0.5+0.3-0.2=0.6$

$\\$

Question 4: If A and B are two events associated with a random experiment such that $\displaystyle P (A \cup B) =0.8, P(A \cap B) = 0.3$ and $\displaystyle P (\overline{A}) = 0.5,$ find $\displaystyle P(B) .$

Answer:

$\displaystyle \text{Given } P (A \cup B) =0.8, P(A \cap B) = 0.3 \text{ and } P (\overline{A}) = 0.5, \text{ find } P(B)$

$\displaystyle P(A) = 1 - P (\overline{A}) = 1 - 0.5 = 0.5$

$\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) \\ \\ \Rightarrow P(B) = P(A \cup B) - P(A) + P(A \cap B) = 0.8-0.5+0.3 = 0.6$

$\\$

Question 5: Given two mutually exclusive events A and B such that $\displaystyle P(A) =\frac{1}{2} \text{ and } P(B)=\frac{1}{3}, \text{ find } P (A \ or \ B).$

Answer:

$\displaystyle \text{Given } P(A) =1/2 \text{ and } P(B)=1/3$

Mutually exclusive events A and B implies $\displaystyle P(A \cap B) = 0$

$\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} =\frac{5}{6}$

$\\$

Question 6: 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.

Answer:

As, out of 3 events A, B and C only one can happen at a time which means no event have anything common.

Therefore We can say that A, B and C are mutually exclusive events.

So, by definition of mutually exclusive events we know that: $\displaystyle P (A \cup B \cup C) = P (A) + P (B) + P (C)$

According to question one event must happen.

So, A or B or C is a sure event.

Therefore $\displaystyle P (A \cup B \cup C) = 1$

We need to find odd against C

$\displaystyle \text{Given: Odd against } A = \frac{8}{3}$

$\displaystyle \frac{P(\overline{A})}{P(A)} = \frac{8}{3}$ $\displaystyle \Rightarrow \frac{1 - P(A) }{P(A)} = \frac{8}{3}$

$\displaystyle 8 P (A) = 3 - 3 P(A) \hspace{0.5cm} \Rightarrow 11 P (A) = 3 \hspace{0.5cm} \Rightarrow P(A) = \frac{3}{11}$

$\displaystyle \text{Given: Odd against } B = \frac{5}{2}$

$\displaystyle \frac{P(\overline{B})}{P(B)} = \frac{5}{2}$ $\displaystyle \Rightarrow \frac{1 - P(B) }{P(B)} = \frac{5}{2}$

$\displaystyle 5 P (B) = 2 - 2 P(B) \hspace{0.5cm} \Rightarrow 7 P (B) = 2 \hspace{0.5cm} \Rightarrow P(B) = \frac{2}{7}$

$\displaystyle P (A \cup B \cup C) = P (A) + P (B) + P (C)$

$\displaystyle \Rightarrow P(C) = P (A \cup B \cup C) - P (A) - P (B) = 1 - \frac{3}{11} - \frac{2}{7} = \frac{34}{77}$

$\displaystyle \text{Therefore, } P(\overline{C}) = 1 - P(C) = 1 - \frac{34}{77} = \frac{43}{77}$

Therefore odd against C are

$\displaystyle \frac{P(\overline{C})}{P(C)} = \frac{\frac{43}{77}}{ \frac{34}{77}} = \frac{43}{34}$

$\\$

Question 7: One of the two events must happen. Given that the chance of one is two-third of the other, find the odds in favor of the other.

Answer:

Let A and B are two events.

As, out of 2 events A and B only one can happen at a time which means no event have anything common.

Therefore we can say that A and B are mutually exclusive events. i.e. $\displaystyle P(A \cap B) = 0$

So, by definition of mutually exclusive events we know that: $\displaystyle P (A \cup B) = P (A) + P (B)$

According to question one event must happen. A or B is a sure event.

So, $\displaystyle P (A \cup B) = P (A) + P (B) = 1$

$\displaystyle \text{Given } P (A) = \frac{2}{3} P (B)$

We have to find the odds in favor of B.

$\displaystyle P (A) + P (B) = 1 \hspace{0.5cm} \Rightarrow \frac{2}{3} P (B) + P(B) = 1 \hspace{0.5cm} \Rightarrow P(B) = \frac{3}{5}$

$\displaystyle \Rightarrow P(\overline{B}) = 1 - P(B) = 1 - \frac{3}{5} = \frac{2}{5}$

Therefore odd against B are

$\displaystyle \frac{P(\overline{B})}{P(B)} = \frac{\frac{3}{5}}{ \frac{2}{5}} = \frac{3}{2}$

$\\$

Question 8: A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of its being a spade or a king.

Answer:

$\displaystyle P(Spade) = \frac{13}{52} = \frac{1}{4}$

$\displaystyle P(King) = \frac{4}{52} = \frac{1}{13}$

$\displaystyle P ( Spade \cap King) = \frac{1}{52}$

$\displaystyle P (Spade \cup King) = P (Spade) + P (King) - P (Spade \ \cap \ King) \\ \\ = \frac{1}{4} + \frac{1}{13} - \frac{1}{52} = \frac{13+4-1}{52} = \frac{4}{13}$

$\\$

Question 9: In a single throw of two dice, find the probability that neither a doublet nor a total of 9 will appear.

Answer:

Let A denotes the event of getting a double.

$\displaystyle \text{So, } A = \{ (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \}$

$\displaystyle \text{Therefore } P(A) = \frac{6}{36} = \frac{1}{6}$

Let B denotes the event of getting a total of 9.

$\displaystyle \text{So, } B = \{ (3,6), (6,3), (4,5), (5,4) \}$

$\displaystyle \text{Therefore } P(B) = \frac{4}{36} = \frac{1}{9}$

We see that A and B are mutually exclusive events. Therefore $\displaystyle P(A \cap B) = 0$

$\displaystyle P( A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{6} + \frac{1}{9} - 0 = \frac{5}{18}$

Therefore the probability of not getting a doublet nor a total of 9 $\displaystyle = 1 - P( A \cup B) = 1 - \frac{5}{18} = \frac{13}{19}$

$\\$

Question 10: A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?

Answer:

We have 1 through 500 integers.

Total number of possible outcomes is $\displaystyle ^{500}C_{1} = 500$

$\displaystyle \text{Therefore, } n (S) = 500$

Integers that are multiples of 3 are $\displaystyle \{ 3, 6, 9, 12, 15, \cdots , 498 \}$

Number of elements $\displaystyle = 166 \ \ \ ( \because 498 = 3 + ( n-1)(3) \Rightarrow n = 165+1 = 166 )$

$\displaystyle P(3) = \frac{166}{500}$

Integers that are multiples of 5 are $\displaystyle \{ 5, 10, 15, 20, 25 \cdots , 500 \}$

Number of elements $\displaystyle = 100 \ \ \ ( \because 500 = 5 + ( n-1)(5) \Rightarrow n = 99+1 = 100 )$

$\displaystyle P(5) = \frac{100}{500}$

Integers that are multiples of 3 and 5 both $\displaystyle \{ 15, 30, 45 \cdots , 495 \}$

Number of elements $\displaystyle = 33 \ \ \ ( \because 495 = 15 + ( n-1)(15) \Rightarrow n = 32+1 = 33)$

$\displaystyle P( 3 \cap 5) = \frac{33}{500}$

$\displaystyle P(3 \cup 5) = P(3) + P(5) - P( 3 \cap 5) = \frac{166}{500} + \frac{100}{500} - \frac{33}{500} = \frac{233}{500}$

$\\$

Question 11: A die is thrown twice. What is the probability that at least one of the two throws come up with the number 3?

Answer:

If a dice is thrown twice, it has a total of $\displaystyle (6 \times 6) = 36$ possible outcomes.

If S represents the sample space then, $\displaystyle n (S) = 36$

Let A represent events the event such that 3 comes in the first throw.

$\displaystyle A = \{(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\}$

$\displaystyle P(A) = \frac{6}{36} = \frac{1}{6}$

Let B represent events the event such that 3 comes in the second throw.

$\displaystyle B = \{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)\}$

$\displaystyle P(A) = \frac{6}{36} = \frac{1}{6}$

It is clear that $\displaystyle (3,3)$ is common in both events so,

$\displaystyle P(A \cap B) = \frac{1}{36}$

$\displaystyle P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{1}{6} =\frac{1}{6} - \frac{1}{36} = \frac{11}{36}$

$\\$

Question 12: A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.

Answer:

$\displaystyle P(Ace) = \frac{4}{52} = \frac{1}{13}$

$\displaystyle P(Spade) = \frac{13}{52} = \frac{1}{4}$

$\displaystyle P ( Ace \cap Spade) = \frac{1}{52}$

$\displaystyle P (Ace \cup Spade) = P (Ace) + P (Spade) - P (Ace \ \cap \ Spade) \\ \\ = \frac{1}{13} + \frac{1}{4} - \frac{1}{52} = \frac{4+13-1}{52} = \frac{4}{13}$

$\\$

Question 13: The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75. what is the probability of passing the Hindi examination.

Answer:

Let E and H denote the events that the student will pass in English and Hindi examination respectively.

$\displaystyle \text{Then, we have } P(English \cap Hindi)=0.5, P(\overline{English} \cap \overline{Hindi} )=0.1 \\ \\ \text{ and } P(English)=0.75$

$\displaystyle P(English \cup Hindi) = 1 - P(\overline{English} \cap \overline{Hindi} ) = 1 - 0.1 = 0.9$

$\displaystyle P(English \cup Hindi) = P(English) + P(Hindi) - P(\overline{English} \cap \overline{Hindi} )$

$\displaystyle \Rightarrow P(Hindi) = P(E \cup Hindi) - P(English) + P(\overline{English} \cap \overline{Hindi} ) \\ \\ = 0.9 - 0.75 +0.5 = 0.65$

$\\$

Question 14: One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

Answer:

We have 1 through 100 integers.

Total number of possible outcomes is $\displaystyle ^{100}C_{1} = 100$

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

Integers that are multiples of 4 are $\displaystyle \{ 4, 8, 12, 16, 20, \cdots , 100 \}$

Number of elements $\displaystyle = 25 \ \ \ ( \because 100 = 4 + ( n-1)(4) \Rightarrow n = 24+1 = 25 )$

$\displaystyle P(4) = \frac{25}{100}$

Integers that are multiples of 6 are $\displaystyle \{ 6, 12, 18, 24, 30 \cdots , 96 \}$

Number of elements $\displaystyle = 16 \ \ \ ( \because 96 = 6 + ( n-1)(6) \Rightarrow n = 15+1 = 16 )$

$\displaystyle P(6) = \frac{16}{100}$

Integers that are multiples of 4 and 6 both $\displaystyle \{ 12, 24, 36, \cdots , 96 \}$

Number of elements $\displaystyle = 8 \ \ \ ( \because 96 = 12 + ( n-1)(12) \Rightarrow n = 7+1 = 8)$

$\displaystyle P( 4 \cap 6) = \frac{8}{100}$

$\displaystyle P(4 \cup 6) = P(4) + P(6) - P( 4 \cap 6) = \frac{25}{100} + \frac{16}{100} - \frac{8}{100} = \frac{33}{100}$

$\\$

Question 15: 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 color.

Answer:

In a deck of 52 cards there are 2 colors. Each color having 26 cards. We need to choose 4 cards out of 52.

Let S represents the sample space. Therefore $\displaystyle n(S) = ^{52}C_{4}$

Let A represents the event that all 4 cards drawn are black.

$\displaystyle n(A) = ^{26}C_{4}$

$\displaystyle P(A) = \frac{^{26}C_{4}}{^{52}C_{4}} = \frac{46}{833}$

Let B represents the event that all 4 cards drawn are red.

$\displaystyle n(B) = ^{26}C_{4}$

$\displaystyle P(B) = \frac{^{26}C_{4}}{^{52}C_{4}} = \frac{46}{833}$

As both events A and B have no common elements or we can say that they are mutually exclusive.

Therefore $\displaystyle P(A \cap B) = 0$

$\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{46}{833} + \frac{46}{833} = \frac{92}{833}$

$\\$

Question 16: 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 passed at least one examination.

Answer:

Total number of student given are 100.

Number of student passed in first examination is 60.

$\displaystyle \text{So, probability of student passing in first examination is, } \\ \\ P(A) = \frac{60}{100} = 0.6$

Number of student passed in second examination is 50.

$\displaystyle \text{So, probability of student passing in second examination is, } \\ \\ P(A) = \frac{50}{100} = 0.5$

Number of student passed in both examination is 30.

$\displaystyle \text{So, probability of student passing in both examination is, } \\ \\ P(A \cap B) = \frac{30}{100} = 0.5$

From probability addition rule,

$\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\displaystyle P(A \cup B) = 0.6 + 0.5 - 0.3$

$\displaystyle P(A \cup B) = 0.8$

Thus, probability of student passing at least in one subject is 0.8.

$\\$

Question 17: A box contains 10 white, 6 red and 10 black balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either white or red?

Answer:

$\displaystyle P(White) = \frac{10}{26} = \frac{5}{13}$

$\displaystyle P(Red) = \frac{6}{26} = \frac{3}{13}$

$\displaystyle P ( White \cap Red) = 0$

$\displaystyle P (White \cup Red) = P (White) + P (Red) - P (White \ \cap \ Red) \\ \\ = \frac{5}{13} + \frac{3}{13} - 0 = \frac{8}{13}$

$\\$

Question 18: In a race, the odds in favor of horses $A, B, C, D \text{ are } 1:3, 1:4, 1:5 \text{ and } 1:6$ respectively. Find probability that one of them wins the race.

Answer:

We have

$\displaystyle P(A):P(\overline{A})=1:3 \hspace{0.2cm} \Rightarrow P(A)=\frac{1}{4}$

$\displaystyle P(B):P(\overline{B})=1:4 \hspace{0.2cm} \Rightarrow P(B)=\frac{1}{5}$

$\displaystyle P(C):P(\overline{C})=1:5 \hspace{0.2cm} \Rightarrow P(C)=\frac{1}{6}$

$\displaystyle P(D):P(\overline{D})=1:6 \hspace{0.2cm} \Rightarrow P(D)=\frac{1}{7}$

Therefore the probability that one of them wins the race is

$\displaystyle P(A \cup B \cup C \cup D )$

$\displaystyle = P(A) + P(B) + P(C) +P(D) = \frac{1}{4} + \frac{1}{5} +\frac{1}{6} +\frac{1}{7} = \frac{105+84+70+60}{420} = \frac{319}{420}$

$\\$

Question 19: The probability that a person will travel by plane is 3/5 and that he will travel by train is 1/4. What is the probability that he (she) will travel by plane or train?

Answer:

$\displaystyle \text{Given } P(T) = \frac{3}{5} \text{ and } P(B) = \frac{1}{4}$

By definition of P(A or B) under axiomatic approach(also called addition theorem)

$\displaystyle \therefore P(T \cup A) = P(T) + P(A) - P(A \cap T)$

A person can never travel by both plane and train simultaneously.

$\displaystyle \therefore P(A \cap T) = 0$

$\displaystyle \text{Hence, } P(T \cup A) = P(T) + P(A) = \frac{3}{5} + \frac{1}{4} = \frac{17}{20}$

$\\$

Question 20: Two cards are drawn from a well shuffled pack of 52 cards. Find the probability that either both are black or both are kings.

Answer:

Let S : Drawing 2 cards out of 52 card $\displaystyle \therefore n(S) = ^{52}C_{2}$

A : Drawing 2 red cards $\displaystyle \therefore n(A) = ^{26}C_{2}$

B : Drawing 2 kings $\displaystyle \therefore n(B) = ^{4}C_{2}$

But there are 2 red kings, so $\displaystyle A \cap B$ : Drawing 2 red kings $\displaystyle \therefore n(A \cap B) = ^{2}C_{2}$

Therefore Required probability $\displaystyle = P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\displaystyle = \frac{n(A)}{n(S)} + \frac{n(B)}{n(S)} -\frac{n(A \cap B)}{n(S)}$

$\displaystyle = \frac{^{26}C_{2}}{^{52}C_{2}} + \frac{^{4}C_{2}}{^{52}C_{2}} -\frac{^{2}C_{2}}{^{52}C_{2}}$

$\displaystyle = \frac{26 \times 25}{52 \times 51}+\frac{4 \times 3}{52 \times 51}- \frac{2}{52 \times 51} = \frac{600}{2652} = \frac{55}{221}$

$\\$

Question 21: In an enhance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examinations 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?

Answer:

Let A: student passes first examination B: student passes second examination

Given: $\displaystyle P(A) = 0.8, P(B) = 0 .7 \text{ and } P(A \cup B) = 0.95$

We need to find $\displaystyle P(A \cap B)$

$\displaystyle \therefore P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.55$

$\\$

Question 22: A box contains 30 bolts and 40 nuts. Half of the bolts and half of the nuts are rusted. If two items are drawn at random, what is the probability that either both are rusted or both are bolts?

Answer:

The number of bolts and nuts are 30 and 40 respectively.

The number of rusted bolts and rusted nuts are 15 and 20 respectively.

Total number of items $\displaystyle =30+40=70$

Total number of rusted items $\displaystyle =15+20=35$

Total number of ways of drawing 2 items $\displaystyle = ^{70}C_{2}$

Let R and B be the events that the both item drawn are rusted items and both are bolts respectively.

R and B are not mutually exclusive events because there are 15 rusted bolts.

$\displaystyle \therefore P ( \text{ items are both rusted or both bolts } ) = P(R \cup B)=P(R)+P(B)-P(R \cap B)$

$\displaystyle = \frac{ ^{35}C_{2} }{ ^{70}C_{2} } + \frac{ ^{30}C_{2} }{ ^{70}C_{2} } - \frac{ ^{15}C_{2} }{ ^{70}C_{2} }$

$\displaystyle = \frac{35 \times 34 }{70 \times 69 } + \frac{30 \times 29 }{70 \times 69 } - \frac{15 \times 14 }{70 \times 69 }$

$\displaystyle = \frac{1850}{4830} = \frac{185}{483}$

$\\$

Question 23: An integer is chosen at random from first 200 positive integers. Find the probability that the integer is divisible by 6 or 8.

Answer:

We have 1 through 200 integers.

Total number of possible outcomes is $\displaystyle ^{200}C_{1} = 200$

$\displaystyle \text{Therefore, } n (S) = 200$

Integers that are multiples of 6 are $\displaystyle \{ 6,12,18, 24 \cdots , 198 \}$

Number of elements $\displaystyle n(A) = 33 \ \ \ ( \because 198 = 6 + ( n-1)(6) \Rightarrow n = 32+1 = 33 )$

Integers that are multiples of 8 are $\displaystyle \{ 8, 16, 24, 32, \cdots , 200 \}$

Number of elements $\displaystyle n(B) = 25 \ \ \ ( \because 200 = 8 + ( n-1)(8) \Rightarrow n = 24+1 = 25 )$

Integers that are multiples of 6 and 8 both $\displaystyle \{ 24, 48, \cdots , 192 \}$

Number of elements $\displaystyle n(A \cap B) = 8 \ \ \ ( \because 192 = 24 + ( n-1)(24) \Rightarrow n = 7+1 = 8)$

$\displaystyle P(A \cup B) = P( \text{ Integer chosen is divisible by 6 or 8 } )$

$\displaystyle = P(A) + P(B) - P(A \cap B)$

$\displaystyle = \frac{n(A)}{n(S)} + \frac{n(b)}{n(S)} -\frac{n(A \cap B)}{n(S)}$

$\displaystyle = \frac{33}{200}+\frac{25}{200}- \frac{8}{200} = \frac{50}{200} = \frac{1}{4}$

$\\$

Question 24: Find the probability of getting 2 or 3 tails when a coin is tossed four times.

Answer:

A coin is tossed four times.

Total number of possible outcomes is $\displaystyle 2^4 = 16$

$\displaystyle \text{Therefore, } n (S) = 16$

Let A be the event of getting 2 tails.

$\displaystyle \text{Therefore } A = 6 \text{ i.e.} \{ TTHH,THTH,THHT,HTTH,HTHT,HHTT \}$

$\displaystyle n (A) = 6$

$\displaystyle P (A) = \frac{n (E_1) }{ n (S)} = \frac{6 }{ 16} = \frac{3}{8}$

Let B be the event of getting 3 tails.

$\displaystyle \text{Therefore } B = 4 \text{ i.e.} \{ TTTH ,TTHT, THTT,HTTT \}$

$\displaystyle n (B) = 4$

$\displaystyle P (B) = \frac{n (E_2) }{ n (S)} = \frac{4 }{ 16} = \frac{1}{4}$

We need to find the probability of getting 2 tails or 3 tails i.e. $\displaystyle P(A \cup B)$

As we can’t get 2 and 3 tails at the same time.

So A and B are mutually exclusive events.

$\displaystyle \therefore P(A \cup B) = P(A) + P(B) = \frac{3}{8} + \frac{1}{4} = \frac{5}{8}$

$\\$

Question 25: Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

Answer:

We have 1 through 1000 integers.

Total number of possible outcomes is $\displaystyle ^{1000}C_{1} = 1000$

$\displaystyle \text{Therefore, } n (S) = 1000$

Integers that are multiples of 2 are $\displaystyle \{ 2, 4, 6, 8, 10, \cdots , 1000 \}$

Number of elements $\displaystyle = 500 \ \ \ ( \because 1000 = 2 + ( n-1)(2) \Rightarrow n = 499+1 = 500 )$

Integers that are multiples of 9 are $\displaystyle \{ 9, 18, 27, 36, \cdots , 999 \}$

Number of elements $\displaystyle = 111 \ \ \ ( \because 999 = 9 + ( n-1)(9) \Rightarrow n = 110+1 = 111 )$

Integers that are multiples of 2 and 9 both $\displaystyle \{ 18, 36, \cdots , 990 \}$

Number of elements $\displaystyle = 55 \ \ \ ( \because 990 = 18 + ( n-1)(18) \Rightarrow n = 54+1 = 55)$

Therefore the number of possible integers $\displaystyle = 500 + 111 - 55 = 556$

$\displaystyle n (E) = 556$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{556 }{ 1000} = 0.556$

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Question 26: ln a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a color television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either any one or both kinds of sets?

Answer:

Let C be the even that family own color television set and B be the event that family owns a black and white television set.

It is given that, $\displaystyle P(C) = 0.87, P(B) = 0.36 \text{ and } P(C \cap B) = 0.30$

We have to find probability that a family owns either anyone or both kind of sets i.e., $\displaystyle P(B \cup C)$

Now, $\displaystyle P(B \cup C) = P(B) + P(C)-P(C \cap B) = 0.87 + 0.36-0.30= 0.93$

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Question 27: If A and B are mutually exclusive events such that $\displaystyle P (A) = 0.35$ and $\displaystyle P (B) = 0.45,$ find

(i) $\displaystyle P (A \cup B)$ (ii) $\displaystyle P (A \cap B)$ (iii) $\displaystyle P (A \cap \overline{B} )$ (iv) $\displaystyle P (\overline{A} \cap \overline{B} )$

Answer:

$\displaystyle P (\overline{A } ) = 1 - P(A) = 1 - 0.35 = 0.65$

$\displaystyle P (\overline{B } ) = 1 - P(B) = 1 - 0.45 = 0.55$

A and B are mutually exclusive events. $\displaystyle P ( A \cap B) = 0$

(i) $\displaystyle P (A \cup B) = P(A)+P(B) - P(A \cap B) = 0.35 +0.45 - 0 = 0.80$

(ii) $\displaystyle P (A \cap B) = 0$

(iii) $\displaystyle P (A \cap \overline{B} ) = P(A) - P(A \cap B) = 0.35 - 0 = 0.35$

(iv) $\displaystyle P (\overline{A} \cap \overline{B} ) = 1 - P(A \cup B) = 1 -0.8 = 0.2$

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Question 28: A sample space consists of 9 elementary event $\displaystyle E_1, E_2,E_3,...,E_8,E_9$ whose probabilities are $\displaystyle P(E_1) = P(E_2) = 0.08, P(E_3) =P(E_4) =0.1, \\ P(E_6) =P(E_7) =0.2, P(E_8) =P(E_9) =0.07$

Suppose $\displaystyle A : \{ E_1, E_5,E_8 \}, \ \ B \{ E_2, E_5, E_8, E_9 \}$

(i) Compute $\displaystyle P (A), P (B)$ and $\displaystyle P (A \cap B)$

(ii) Using the addition law of probability, find $\displaystyle P (A \cup B)$

(iii) List the composition of the event $\displaystyle A \cup B,$ and calculate $\displaystyle P(A \cup B)$ by adding the probabilities of the elementary events

(iv) Calculate $\displaystyle P (\overline{B})$ from $\displaystyle P(B),$ also calculate $\displaystyle P (\text{B})$ directly from the elementary events of $\displaystyle \overline{B}.$