Question

An officer has 3 options to go to his offices (i) By bike (ii) by car (iii) by bus. The probabilities of going late to his office in these three options are $\frac { 1 } { 4 } , \frac { 1 } { 3 } , \frac { 1 } { 2 }$ respectively. Every day he tosses a die. If an even prime is thrown, he goes by bike. If odd prime is thrown, he goes by car or else, he goes by bus. If he is late his office, on a particular day, then the probability that he traveled by car on that day

• A. 24/87
• B. 48/87
• C. 36/87
• D. 8/87

Answer 1

Answer: (A)

24/87

Answer 2

Let A, B, C be the events of choosing bike, car and bus

respectively.

P(1) = $\frac { 1 } { 2 }$

E be the event of going late to his office

P(E/A) = $\frac { 1 } { 4 }$, P(E/B) = $\frac { 1 } { 3 }$ ; P(E/C) = $\frac { 1 } { 2 }$

P(B/E) = $\frac { \mathrm { P } ( \mathrm { B } ) \cdot \mathrm { P } ( \mathrm { E } / \mathrm { B } ) } { \mathrm { P } ( \mathrm { A } ) \cdot \mathrm { P } ( \mathrm { E } / \mathrm { A } ) + \mathrm { P } ( \mathrm { B } ) \cdot \mathrm { P } ( \mathrm { E } / \mathrm { B } ) + \mathrm { P } ( \mathrm { C } ) \mathrm { P } ( \mathrm { E } / \mathrm { C } ) }$

= $\frac { \frac { 1 } { 3 } \times \frac { 1 } { 3 } } { \frac { 1 } { 24 } + \frac { 1 } { 9 } + \frac { 1 } { 4 } } = \frac { 1 } { 9 } \times \frac { 1 } { \frac { 7 } { 24 } + \frac { 1 } { 9 } }$

= $\frac { 1 } { 9 } \times \frac { 24 \times 9 } { 63 + 24 } = \frac { 24 } { 87 } + \frac { 8 } { 29 }$.