Question

. A box contains b blue balls and y yellows balls. We take out a ball randomly at a time repeatedly. A ball which are taken out does not go back into the box. (a) Compute the probability that the first yellow ball drawn is the (k +1)th ball. (Hint: The computation may be simpler if you use Problem 3 (b).) (b) Compute the probability that the last ball drawn is yellow.

Answer 1

a)

first yellow ball is (k+1)th ball

hence

first k balls are blue and (k+1)th is yellow

required number of ways =

$\binom{b}{k}\binom{y}{1}$

total number of ways to select (k+1) balls without any contraint

$= \binom{b+y}{k+1}$

hence

required probability =

$= \frac{\binom{b}{k}\binom{y}{1}}{ \binom{b+y}{k+1}}$

b)

last ball is yellow means that in (b+y -1) balls , there are b blue balls and y-1 yellow balls

number of ways for same condition

$=\binom{b+y-1}{b}$

total number of ways without any constraint

$= \binom{b+y}{b}$

hence

required probability =

$= \frac{\binom{b+y-1}{b}}{\binom{b+y}{b}}$