Question

A bag contains 14 identical balls of which 5 are red, 3 black and 6 white balls. Four balls are drawn at random from the bag. Find the probability that (i) one is white

Answer 1

Out of $14$ balls $4$ balls can be drawn in $^{14}C_4=1001$ ways.

Now one white ball can be drawn in such ways

(1) $1$ white $3$ red balls

(2) $1$ white $3$ black balls

(3) $1$ white $1$ red $2$ black balls

(4) $1$ white $2$ red $1$ black balls

Therefore, $1$ white $3$ red balls can be drawn in $( ^6C_1× ^{5}C_3)$$=10$ ways.

$1$ white $3$ black balls can be drawn in $( ^6C_1× ^{3}C_3)$$=6$ ways.

$1$ white $1$ red $2$ black balls can be drawn in $(^6C_1× ^{5}C_1× ^{3}C_2)$= $90$ ways.

$1$ white $2$ red $1$ black balls can be drawn in $(^6C_1× ^{5}C_2× ^{3}C_1)$ $=180$ ways.

Let $A$ be an event such that out of four drawn balls one ball is white.

Then number of elements in favour $A$ is $=(10+6+90+180)=286$

Therefore, $P(A)=\frac{n(A)}{n(S)}=\frac{286}{1001}=\frac{2}{7}$