A problem by Ankit Nigam

Level pending

If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, and ball is drawn at random, then the probability that 2 white ball and 1 black will be drawn is

PRACTICE FOR BITSAT HERE

3/16 13/32 1/4 1/32

1 solution

Ankit Nigam
Apr 22, 2015

$In\quad a\quad simple\quad way$

$Case\quad 1:\quad WWB\\ { p }_{ 1 }=\frac { 3 }{ 4 } \times \frac { 2 }{ 4 } \times \frac { 3 }{ 4 }$

$Case\quad 2:\quad WBW\\ { p }_{ 2 }=\frac { 3 }{ 4 } \times \frac { 2 }{ 4 } \times \frac { 1 }{ 4 }$

$Case\quad 3:\quad BWW\\ { p }_{ 3 }=\frac { 1 }{ 4 } \times \frac { 2 }{ 4 } \times \frac { 1 }{ 4 }$

$\therefore p={ p }_{ 1 }+{ p }_{ 2 }+{ p }_{ 3 }\\ =\frac { 26 }{ 64 } \\ \boxed { =\frac { 13 }{ 32 } }$

A problem by Ankit Nigam

1 solution

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