Question

Two persons A and B throw a die alternately till one of them

get a “six” and wins the game. The probability of winning of

B is-

• A. $\frac { 6 } { 11 }$
• B. $\frac { 5 } { 11 }$
• C. $\frac { 3 } { 11 }$
• D. None of these

Answer 1

Answer: (B)

$\frac { 5 } { 11 }$

Answer 2

Let E = the event that A gets six, P(5) = $\frac { 1 } { 6 }$

F = the event that B gets six, P(F) = $\frac { 1 } { 6 }$

\ P(B wins) = P($\overline { \mathrm { E } }$F or $\overline { \mathrm { E } }$ $\overline { \mathrm { E } }$ F or $\overline { \mathrm { E } }$ $\overline { \mathrm { E } }$ $\overline { \mathrm { E } }$ F …….)

(Since B can win the game in 2^nd, 4^th, 6^th …… throw)

= $\left( \frac { 5 } { 6 } \right) ^ { 3 }$ $\frac { 1 } { 6 }$ + $\left( \frac { 5 } { 6 } \right) ^ { 5 }$ $\frac { 1 } { 6 }$ + ………. = $\frac { 5 } { 36 }$

= $\frac { 5 } { 36 }$ $\frac { 1 } { 1 - \frac { 25 } { 36 } }$ = $\frac { 5 } { 11 }$