Question

A student takes his examination in four subjects : $S_1,S_2,S_3$ and $S_4$. His chances of passing in these are $\frac{4}{5}$, $\frac{3}{4}$, $\frac{5}{6}$, and $\frac{2}{3}$ respectively. To qualify he must pass in $S_1$ and in at least two others. What is the probability that he qualifies ?

• A. $\frac{61}{90}$
• B. $\frac{63}{90}$
• C. $\frac{29}{90}$
• D. None of these
  

Answer 1

The correct option is A $\frac{61}{90}$

$P\left(S_1\right)=\frac{4}{5},P\left(S_2\right)=\frac{3}{4},P\left(S_3\right)=\frac{5}{6},P\left(E_4\right)=\frac{2}{3}$

i) Probability of passing in $S_2,S_3$ and failing $S_4$
$=\frac{3}{4}\times \frac{5}{6}\times (1-\frac{2}{3})$
$=\frac{3}{4}\times \frac{5}{6}\times \frac{1}{3}=\frac{5}{24}$

ii) Probability of posing in $S_3{S}_4$ and failing $S_2$
$=\frac{5}{6}\times \frac{2}{3}\times (1-\frac{3}{4})=\frac{5}{6}\times \frac{2}{3}\times \frac{1}{4}=\frac{5}{36}$

iii) Probability of passing $S_2{S}_4$ and failing $S_3$
$=\frac{2}{3}\times \frac{3}{4}\times (1-\frac{5}{6})=\frac{2}{3}\times \frac{3}{4}\times \frac{1}{6}=\frac{1}{12}$

iv) Probability of passing $S_2{S}_3{S}_4$
$=\frac{3}{4}\times \frac{5}{6}\times \frac{2}{3}=\frac{5}{12}$
Hence the probability of passing at least two subjects
$=\frac{5}{24}+\frac{5}{36}+\frac{1}{12}+\frac{5}{12}=\frac{61}{72}$
Also probability of passing $S_1$ and at least two subject
$=\frac{4}{5}\times \frac{61}{72}=\frac{61}{90}$