Question

Only four students A, B, C, D appear in an examination. The probability of A coming first is 2 times that of B, probability of B coming first is 3 times that of C and probability of C coming first is four times that of D. The probability that either A or D comes first is

• A. 24/41
• B. 25/41
• C. 16/41
• D. 17/41

Answer 1

Answer: (B)

25/41

Answer 2

Given, P(1) = 2P(2); P(2) = 3P(3) and P(3) = 4P(4)

Clearly P(2) =

P(3) =

P(4) = $\frac { \mathrm { P } ( \mathrm { C } ) } { 4 } = \frac { \mathrm { P } ( \mathrm { A } ) } { 24 }$

Since there are only four students, hence events of ‘coming first’ is mutually exclusive.

∴ P(1) + P(2) + P(3) + P(4) = 1

or, P(1) $\left( 1 + \frac { 1 } { 2 } + \frac { 1 } { 6 } + \frac { 1 } { 24 } \right) = 1$

⇒ P(1) = $\frac { 24 } { 41 }$

∴ P(AUD) = P(1) + P(4)

= P(1) + $\frac { \mathrm { P } ( \mathrm { A } ) } { 24 } = \frac { 25 } { 24 } \mathrm { P } ( \mathrm { A } )$

= $\frac { 24 } { 41 } \times \frac { 25 } { 24 } = \frac { 25 } { 41 }$