Question

Let p denotes the probability that a managed x years will die in a year. The probability that out of n men each aged x, $A _ { 1 }$ will die in a year and will be the first to die, is

• A. $\frac { 1 } { n } \left[ 1 - ( 1 - p ) ^ { n } \right]$
• B. $\left[ 1 - ( 1 - p ) ^ { n } \right]$
• C. $\frac { 1 } { n - 1 } \left[ 1 - ( 1 - p ) ^ { n } \right]$
• D. None of these

Answer 1

Answer: (A)

$\frac { 1 } { n } \left[ 1 - ( 1 - p ) ^ { n } \right]$

Answer 2

Let E_i denotes the event that A_i dies in a year.

Then $P \left( E _ { i } \right) = p$ and $P \left( E _ { i } ^ { \prime } \right) = 1 - p$ for i = 1, 2, ….n

P (none of $A _ { 1 } , A _ { 2 } , \ldots . . A _ { 3 }$ dies in a year)

$= P \left( E _ { 1 } ^ { \prime } \cap E _ { 2 } ^ { \prime } \cap \ldots . . E _ { n } ^ { \prime } \right) = P \left( E _ { 1 } ^ { \prime } \right) P \left( E _ { 2 } ^ { \prime } \right) \ldots P \left( E _ { n } ^ { \prime } \right) = ( 1 - p ) ^ { n }$ ,

because are independent.

Let E denote the event that at least one of dies in a year.

Then

Let F denote the event that $A _ { 1 }$ is the first to die.

Then $P ( F / E ) = \frac { 1 } { n }$ . Also, $P ( F ) = P ( E ) \cdot P ( F / E ) = \frac { 1 } { n } \left[ 1 - ( 1 - p ) ^ { n } \right]$ .