Question

A bag has $4$ red and $5$ black balls and a second bag has $3$ red and $7$ black balls. One ball is drawn form the first and two from the second. Find the probability that out of three balls two are black and one is red.

Answer 1

In first bag there are $4$ red and $5$ black balls and in second bag there are $3$ red and $7$ black balls.

P( two red and one black ball )

$=$ P ( One red from first bag and one red one black from second bag or one black ball from first bag and two red balls from second bag )

$=$ P ( one red from first bag and one red one black from second bag ) + P (one black ball from first bag and two red balls from second bag )

$=\frac{4}{9}\times \frac{{{}^{3}C}_1\times {{}^{7}C}_1}{{{}^{10}C}_2}+\frac{5}{9}\times \frac{{{}^{3}C}_2}{{{}^{10}C}_2}$

$=\frac{4}{9}\times \frac{\frac{3}{1}\times \frac{7}{1}}{\frac{10\times 9}{1\times 2}}+\frac{5}{9}\times \frac{\frac{3\times 2}{1\times 2}}{\frac{10\times 9}{1\times 2}}$

$=\frac{4}{9}\times \frac{7}{15}+\frac{5}{9}\times \frac{3}{15}=\frac{28}{135}+\frac{5}{135}$

$=\frac{33}{135}$

$=\frac{11}{45}.$

Hence, the answer is $\frac{11}{45}.$