Question

An urn contains $5$ - red and $6$ - blue balls. Two balls are drawn at random with replacement. Find the probability of getting
(1) both red balls
(2) both blue balls
(3) one-red and one-blue ball

Answer 1

total number of balls are $11(5-Red+6-blue)$

therefore the probability of getting both red balls is =$\frac{5}{11}\times \frac{5}{11}=\frac{25}{121}$

similarly probability of getting both blue balls = $\frac{6}{11}\times \frac{6}{11}=\frac{36}{121}$

similarly probability of getting $1$ red and $1$ green is =$\frac{5}{11}\times \frac{6}{11}=\frac{30}{121}$