A candidate has to go to the examination centre to appear in an examination.

Question

A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter a...

Accepted Answer

$\begin{gathered} \quad  \quad  \quad  \quad  \quad  \quad  \ \frac{2}{5} \quad  \frac{1}{5} \quad  \frac{2}{5} \quad  \quad  \frac{1}{5} \quad  \frac{1}{3} \quad  \frac{1}{4} \end{gathered}$

$\begin{gathered} \begin{gathered} \quad \quad \quad \quad \quad \quad \quad \ B \quad S \quad C \quad \quad {L}_{b} \quad {L}_{s} \quad {L}_{c} \end{gathered} \end{gathered}$

We label the event that the candidate arrives late be $L$

$\begin{gathered} B \quad  S \quad  C \quad  {L}_{b} \quad  {L}_{s} \quad  {L}_{c} \ \frac{2}{5} \quad  \frac{1}{5} \quad  \frac{2}{5} \quad  \frac{1}{5} \quad  \frac{1}{3} \quad  \frac{1}{4} \end{gathered}$

$\Rightarrow P(L\cap B)=P(L)\cdot P(B/L)=P(B\cap L)$

$\Rightarrow$ $P(B/L)=\frac{P(B\cap L)}{P(L)}$

$=\frac{P(B)\cdot P(L/B)}{P(B\cap L)+P(S\cap L)+P(C\cap L)}$

$=\frac{P(B)\cdot P(L/B)}{P(B)\cdot P(L/B)+P(S)\cdot P(L/S)+P(C)\cdot P(C/B)}$

$P(B/L)=\frac{P(B)\cdot P({L}_{b})}{P(B)\cdot P({L}_{b})+P(S)\cdot P({L}_{s})+P(C)\cdot P({L}_{c})}$

$=\frac{(\frac{2}{5})\cdot (\frac{1}{5})}{(\frac{2}{5})\cdot (\frac{1}{5})+(\frac{1}{5})\cdot (\frac{1}{3})+(\frac{2}{5})\cdot (\frac{1}{4})}$

$=(\frac{2}{25})\cdot (\frac{300}{24+20+30})=\frac{12}{37}$

Solutions · 1