Question

Two events A and B such that $P\left( A' \right)=0.3,P\left( B \right)=0.5$ and $P\left( A\cap B \right)=0.3$, then $P\left( B|A\cup B' \right)$ is:
(a) $\dfrac{3}{8}$
(b) $\dfrac{2}{3}$
(c) $\dfrac{5}{6}$
(d) $\dfrac{1}{4}$

Answer 1

Hint: In this question, we first need to look into the addition theorem of probability. We need to solve it using the conditional probability and then substitute the equations from the addition theorem of probability in order to simplify further.

Complete step-by-step answer:
$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$

& P\left( A\cap B' \right)=P\left( A \right)-P\left( A\cap B \right) \\\ & P\left( B|A \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)} \\\ \end{aligned}$$ $$P\left( A \right)+P\left( A' \right)=1$$ $$P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$$ Let us look into some basic definitions at first. TRIAL: Let a random experiment be repeated under identical conditions then the experiment is called a trial. OUTCOME: A possible result of a random experiment is called its outcome. SAMPLE SPACE: The set of all possible outcomes of an experiment is called the sample space of the experiment and is denoted by S. SAMPLE POINT: The outcome of an experiment is called sample point. EVENT: A subset of the sample space associated with a random experiment is said to occur, if any one of the elementary events associated to it is an outcome. PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by $$P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$$ Addition Theorem of Probability: For two events A and B, $$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$$ If A and B are two events associated with a random experiment, then $$P\left( A\cap B' \right)=P\left( A \right)-P\left( A\cap B \right)$$ CONDITIONAL PROBABILITY: Let A and B be two events associated with a random experiment. Then, the probability of occurrence of event B under the condition that A has already occurred and $$P\left( A \right)\ne 0$$, is called the conditional probability. $$P\left( B|A \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$$ Now, from the given conditions in the question we get, $$P\left( A' \right)=0.3,P\left( B \right)=0.5$$ $$P\left( A\cap B \right)=0.3$$ $$\begin{aligned} & P\left( A \right)=1-P\left( A' \right) \\\ & P\left( A \right)=1-0.3 \\\ & \therefore P\left( A \right)=0.7 \\\ \end{aligned}$$ $$\begin{aligned} & P\left( B' \right)=1-P\left( B \right) \\\ & P\left( B' \right)=1-0.5 \\\ & \therefore P\left( B' \right)=0.5 \\\ \end{aligned}$$ Now, by applying the conditional probability formula we get, $$\begin{aligned} & \Rightarrow P\left( B|A\cup B' \right)=\dfrac{P\left( B\cap \left( A\cup B' \right) \right)}{P\left( A\cup B' \right)} \\\ & \left[ \because B\cap \left( A\cup B' \right)=\left( B\cap A \right)\cup \left( B\cap B' \right) \right] \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow P\left( B|A\cup B' \right)=\dfrac{P\left[ \left( B\cap A \right)\cup \left( B\cap B' \right) \right]}{P\left( A \right)+P\left( B' \right)-P\left( A\cap B' \right)} \\\ & \left[ \because \left( B\cap B' \right)=\varphi \right] \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow P\left( B|A\cup B' \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)+P\left( B' \right)-P\left( A \right)+P\left( A\cap B \right)} \\\ & \left[ \because P\left( A\cap B' \right)=P\left( A \right)-P\left( A\cap B \right) \right] \\\ \end{aligned}$$ $$\Rightarrow P\left( B|A\cup B' \right)=\dfrac{P\left( A\cap B \right)}{P\left( B' \right)+P\left( A\cap B \right)}$$ $$\begin{aligned} & \Rightarrow P\left( B|A\cup B' \right)=\dfrac{0.3}{0.3+0.5} \\\ & \left[ \because P\left( A\cap B \right)=0.3,P\left( B' \right)=0.5 \right] \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow P\left( B|A\cup B' \right)=\dfrac{0.3}{0.8} \\\ & \therefore P\left( B|A\cup B' \right)=\dfrac{3}{8} \\\ \end{aligned}$$ Hence, the correct option is (a). Note: It is important to consider the complement function wherever given because neglecting any of the complement functions or interchanging it with other functions completely changes the result. $$P\left( A\cap B' \right)=P\left( A \right)-P\left( A\cap B \right)$$ $$P\left( A'\cap B \right)=P\left( B \right)-P\left( A\cap B \right)$$ We need to be careful while applying the additional theorem of probability and conditional theorem of probability because substituting or considering one term in place of the other term completely changes the equation and then the result.