Question
The probability that a student passes at least one of the three examinations $A,B,C$ is $\frac{3}{4}$. The probability that he passes atleast two of them is $\frac{1}{2}$ and the probability he passes exactly two of them is $\frac{2}{5}$. If $a,b,c$ are the probabilities of the student passing in $A,B,C$ respectively, then
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Solution
Sol. [Key Concept: Inclusion–exclusion on the pass events from the given intersection data.]
$A \rightarrow$ event that candidate passes in exam A .
B → event that candidate passes in exam B.
$C \rightarrow$ event that candidate passes in exam C .
$$\begin{matrix} P(\text{ }A \cap \text{ }B \cap C) & \ = \frac{1}{2}\frac{2}{5} = \frac{1}{10} = abc \\ \ \frac{2}{5} & \ \ = \sum_{}^{}\ P(\text{ }A \cap \text{ }B) - 3P(\text{ }A \cap \text{ }B \cap C) \\ \Rightarrow \ \sum_{}^{}P(\text{ }A \cap \text{ }B) & \ = \frac{7}{10} \end{matrix}$$
$$\begin{matrix} \frac{3}{4} & \ = \sum_{}^{}\ P(\text{ }A) - \sum_{}^{}\ P(\text{ }A \cap \text{ }B) + P(\text{ }A \cap \text{ }B \cap C) \\ \Rightarrow & \ \sum_{}^{}P(\text{ }A) = \frac{27}{20} = a + b + c \end{matrix}$$
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