Question

An urn contains 2 white and 2 blacks balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. Find the probability that the third ball drawn is black.

• A. $\frac{17}{30}$
• B. $\frac{16}{23}$
• C. $\frac{7}{15}$
• D. $\frac{23}{30}$

Answer 1

The correct option is D $\frac{23}{30}$

Let $B_i$ = ith ball drawn is black.
$W_i$ = ith ball drawn is white, where i = 1, 2
and A = third ball drawn is black.
We observe that the black ball can be drawn in the third draw in one of the following mutually exclusive ways.
(i) Both first and second balls drawn are white and the third ball drawn is black.
i.e. $(W_1\cap W_2)\cap A$
(ii) Both first and second balls are black and third ball drawn is black.
i.e. $(B_1\cap B_2)\cap A$
(iii) The first ball drawn is white, the second ball drawn is black and the third ball drawn is black.
i.e $(W_1\cap B_2)\cap A$
(iv) The first ball drawn is black, the second ball drawn is white and the third ball drawn is black.
i.e. $(B_1\cap W_2)\cap A$
$\therefore P\left(A\right)=P\left[\right\{(W_1\cap W_2)\cap A\}\cup \{(B_1\cap B_2)\cap A\}$
$\cup \left\{\right(W_1\cap B_2)\cap A\}\cup \left\{\right(B_1\cap W_2)\cap A\}]$
$=P\left\{\right(W_1\cap W_2)\cap A\}+P\left\{\right(B_1\cap B_2)\cap A\}$
$+P\left\{\right(W_1\cap B_2)\cap A\}+P\left\{\right(B_1\cap W_2)\cap A\}$
$=P(W_1\cap W_2).P\left(A|(W_1\cap W_2)\right)+P(B_1\cap B_2).P\left(A|(B_1\cap B_2)\right)$
$+P(W_1\cap B_2).P\left(A|(W_1\cap B_2)\right)+P(B_1\cap W_2).P\left(A|(B_1\cap W_2)\right)$
$=(\frac{2}{4}\times \frac{1}{3})\times 1+(\frac{2}{4}\times \frac{3}{5})\times \frac{4}{6}+(\frac{2}{4}\times \frac{2}{3})\times \frac{3}{4}+(\frac{2}{4}\times \frac{2}{5})\times \frac{3}{4}$
$=\frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{3}{20}=\frac{23}{30}$