Question

A student takes his examination in four subjects $\alpha ,\beta ,\gamma ,\delta$. He estimates his chance of passing in $\alpha$ is $\frac{4}{5}$, in $\beta$ is $\frac{3}{4}$, in $\gamma$ is $\frac{5}{6}$ and in $\delta$ is $\frac{2}{3}$. The probability that he qualifies (passes in atleast three subjects) is

• A. $\frac{34}{90}$
• B. $\frac{61}{90}$
• C. $\frac{53}{90}$
• D. None of these

Answer 1

Answer: (D)

None of these

$P\left(\alpha \right)=\frac{4}{5},P\left(\overline{\alpha }\right)=1-\frac{4}{5}=\frac{1}{5}$
$P\left(\beta \right)=\frac{3}{4},P\left(\overline{\beta }\right)=1-\frac{3}{4}=\frac{1}{4}$
$P\left(\gamma \right)=\frac{5}{6},P\left(\overline{\gamma }\right)=1-\frac{5}{6}=\frac{1}{4}$
$P\left(\delta \right)=\frac{2}{3},P\left(\overline{\delta }\right)=1-\frac{2}{3}=\frac{1}{3}.$

Different possibilities to qualify are
(1) passes in $\alpha ,\beta ,\gamma$ and fails in $\delta$
(2) passes in $\alpha ,\beta ,\delta$ and fails in $\gamma$
(3) passes in $\alpha ,\gamma ,\delta$ and fails in $\beta$

(4) passes in $\beta ,\gamma ,\delta$ and fails in $\alpha$
(5) passes in all the four sujbects $\alpha ,\beta ,\gamma$ and $\delta .$
These are mutually exclusive possibilities.
$\therefore$ Required probability
$=(\frac{4}{5}\times \frac{3}{4}\times \frac{5}{6}\times \frac{1}{3})+(\frac{4}{5}\times \frac{3}{4}\times \frac{2}{3}\times \frac{1}{6})+(\frac{4}{5}\times \frac{5}{6}\times \frac{2}{3}\times \frac{1}{4})+(\frac{3}{4}\times \frac{5}{6}\times \frac{2}{3}\times \frac{1}{5})+(\frac{4}{5}\times \frac{3}{4}\times \frac{5}{6}\times \frac{2}{3})$
$=\frac{1}{6}+\frac{1}{15}+\frac{1}{9}+\frac{1}{12}+\frac{1}{3}=\frac{30+12+20+15+60}{180}=\frac{137}{180}$