Chapter 11, Problem 29MCQ

Solution

a)

To explain: Whether the events A, and B are mutually exclusive or not and also verify if they are independent on not.

Events A and B are not mutually exclusive events and they are independent events.

Given:The set of numbers $=\left\{90,91,92,93,94,95,,96,97,98,99\right\}$ .

Event A is selecting a multiple of 2 and event B is selecting a multiple of 3.

Concept used:

Two events A and B are said to be mutually exclusive when they do not occur simultaneously and $P(A\cap B)=0$ for mutually exclusive events. Two events A and b are said to be independent events when the occurrence of one event does not impact the occurrence of the other event and $P(A\cap B)=P(A)\times P(B)$ holds for two independent events A, B.

Calculation:

With respect to the given events, the probability of event A is calculated as shown below

The numbers in the set which are multiples of 2 are $=\left\{90,92,94,96,98\right\}$ .

The total number of numbers in the set $=10$ .

The probability that randomly selected number is a multiple of 2 is $P(A)=\frac{5}{10}$ .

Similarly the probability that event B is calculated as shown below

The numbers in the set which are multiples of 3 are $=\left\{90,93,96,99\right\}$ .

The total number of numbers in the set $=10$ .

The probability that a randomly selected number is a multiple of 3 is $P(B)=\frac{4}{10}$ .

The numbers which are multiples of both 2 and 3 are $=\left\{90,96\right\}$

The probability that a randomly selected number is a multiple of both 2 and 3 is $P(A\cap B)=\frac{2}{10}$ .

This satisfy the identity

  $\begin{array}{c}P(A\cap B)=P(A)\times P(B)\\ =\frac{5}{10}\times \frac{4}{10}\\ =\frac{2}{10}\end{array}$

Conclusion:The events A and B can occur simultaneously and $P(A\cap B)=\frac{2}{10}\ne 0$ . Thus it can be said that events A and B are not mutually exclusive and it satisfy the identity $P(A\cap B)=P(A)\times P(B)$ , thus it can be said that both events A, B are independent.

b)

To calculate: The probabilities of occurrence of events A and B.

The probabilities of occurrence of events A and B are $P(A)=0.5P(B)=0.4$

Concept used:

The probabilities of a particular event is the ratio of number of favourable events to the total possible outcomes.

Calculation:With respect to the given events, the probability of event A is calculated as shown below

The numbers in the set which are multiples of 2 are $=\left\{90,92,94,96,98\right\}$ .

The total number of numbers in the set $=10$ .

The probability that a randomly selected number is a multiple of 2 is $P(A)=\frac{5}{10}=0.5$ .

Similarly, the probability that event B is calculated as shown below

The numbers in the set which are multiples of 3 are $=\left\{90,93,96,99\right\}$ .

The total number of numbers in the set $=10$ .

The probability that a randomly selected number is a multiple of 3 is $P(B)=\frac{4}{10}=0.4$ .

Conclusion:The probabilities of the occurrence of the events A and B are

  $P(A)=0.5P(B)=0.4$

c)

To calculate: The probability that both events A and B occur together.

The probability that both events A and B occur together is$P(A\cap B)=0.2$.

Concept used:

Two events A and B are said to be independent events when the occurrence of one event does not impact the occurrence of the other event and $P(A\cap B)=P(A)\times P(B)$ holds for two independent events A, B.

Calculation:

The given events A and B are independent events and its probability is calculated as shown below

  $\begin{array}{c}P(A\cap B)=P(A)\times P(B)\\ =\frac{5}{10}\times \frac{4}{10}\\ =\frac{2}{10}\end{array}$

It can also be calculated as shown below

The numbers which are multiples of both 2 and 3 are $=\left\{90,96\right\}$

The probability that a randomly selected number is a multiple of both 2 and 3 is $P(A\cap B)=\frac{2}{10}=0.2$ .

Conclusion:The probability that events A and B occur simultaneously is $P(A\cap B)=\frac{2}{10}$.

d)

To calculate: The probability that both events A or B occur $P(A\cup B)$ .

The probability that both events A or B occur $P(A\cup B)=0.7$

Formula used:

The probability that events A or B is calculated using the formula

  $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Calculation:

The probability that events A or B occur is calculated as shown below

  $\begin{array}{l}P(A)=0.5\\ P(B)=0.4\\ P(A\cap B)=0.2\end{array}$

  $\begin{array}{c}P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ =0.5+0.4-0.2\\ =0.7\end{array}$

Conclusion:The probability that both events A or B occur $P(A\cup B)=0.7$ .

e)

To calculate: The conditional probabilities $P(A|B)\text{ and }P(B|A)$ .

The conditional probabilities $P(A|B)=0.5\text{ and }P(B|A)=0.4$.

Formula used:

The conditional probabilities are calculated using the formula

  $\begin{array}{l}P(A|B)=\frac{P(A\cap B)}{P(B)}\\ P(B|A)=\frac{P(A\cap B)}{P(A)}\end{array}$

Calculation:

The conditional probabilities are calculated as shown below

  $\begin{array}{l}P(A)=0.5\\ P(B)=0.4\\ P(A\cap B)=0.2\end{array}$

  $\begin{array}{c}P(A|B)=\frac{P(A\cap B)}{P(B)}\\ =\frac{0.2}{0.4}\\ =0.5\\ P(B|A)=\frac{P(A\cap B)}{P(A)}\\ =\frac{0.2}{0.5}\\ =0.4\end{array}$

Conclusion:The events A and B are independent so the occurrence of one event has no impact on other event. The conditional probabilities are

  $\begin{array}{l}P(A|B)=P(A)=0.5\\ P(B|A)=P(B)=0.4\end{array}$