Question

A bag contains 5 balls of unknown colors, two balls are drawn at random from the bag and are found to be red. The probability that the bag contains exactly 4 red balls is.

• A. $\frac{3}{5}$
• B. $\frac{3}{10}$
• C. $\frac{5}{13}$
• D. $\frac{4}{13}$

Answer 1

The correct option is B $\frac{3}{10}$

There are 5 balls, 2 of which are given to be red.

So,the bag may contain 2 or 3 or 4 or 5 red balls.

E1: 2 red balls

E2: 3 Red balls

E3: 4 Red balls

E4: 5 Red balls

A: 2 balls are red

$P\left(\frac{E3}{A}\right)$ is required.

$P\left(\frac{E3}{A}\right)=\frac{P\left(\frac{A}{E3}\right).P\left(E3\right)}{P\left(\frac{A}{E1}\right).P\left(E1\right)+P\left(\frac{A}{E2}\right).P\left(E2\right)+P\left(\frac{A}{E3}\right).P\left(E3\right)+P\left(\frac{A}{E4}\right).P\left(E4\right)}$

All E1,E2,E3,E4 are equally likely. so

$P\left(E1\right)=P\left(E2\right)=P\left(E3\right)=P\left(E4\right)=\frac{1}{4}$

$P\left(\frac{A}{E1}\right)=\frac{{}_2^{2}C}{{}_2^{5}C}=\frac{1}{10}P\left(\frac{A}{E2}\right)=\frac{{}_2^{3}C}{{}_2^{5}C}=\frac{3}{10}P\left(\frac{A}{E3}\right)=\frac{{}_2^{4}C}{{}_2^{5}C}=\frac{6}{10}P\left(\frac{A}{E4}\right)=\frac{{}_2^{5}C}{{}_2^{5}C}=1$

So,the probability that the bag contains 4 red balls

$=\frac{\frac{6}{10}.\frac{1}{4}}{\frac{1}{10}.\frac{1}{4}+\frac{3}{10}.\frac{1}{4}+\frac{6}{10}.\frac{1}{4}+1.\frac{1}{4}}=\frac{3}{10}$