- Home
- Standard 11
- Mathematics
Urn $A$ contains $6$ red and $4$ black balls and urn $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$, the probability that it is found to be red, is
$\frac{{32}}{{55}}$
$\frac{{21}}{{55}}$
$\frac{{19}}{{55}}$
None of these
Solution
(a) Let the events are
${R_1} = A$ red ball is drawn from urn $A$ and placed in $B$
${B_1} = A$ black ball is drawn from urn $A$ and placed in $B$
${R_2} = A$ red ball is drawn from urn $B$ and placed in $A$
${B_2} = A$ black ball is drawn from urn $B$ and placed in $A$
$R = A$ red ball is drawn in the second attempt from $A$
Then the required probability
$ = P({R_1}{R_2}R) + ({R_1}{B_2}R) + P({B_1}{R_2}R) + P({B_1}{B_2}R)$
$ = P({R_1})P({R_2})P(R) + P({R_1})P({B_2})P(R) + P({B_1})P({R_2})P(R) + $
$P({B_1})P({B_2})P(R)$
$ = \frac{6}{{10}} \times \frac{5}{{11}} \times \frac{6}{{10}} + \frac{6}{{10}} \times \frac{6}{{11}} \times \frac{5}{{10}} + \frac{4}{{10}} \times \frac{4}{{11}} \times \frac{7}{{10}} + \frac{4}{{10}} \times \frac{7}{{11}} \times \frac{6}{{10}}$
$ = \frac{{32}}{{55}}$.
