Question

A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green.

• A. $\frac{23}{40}$
• B. $\frac{21}{40}$
• C. $\frac{19}{40}$
• D. $\frac{17}{40}$

Answer 1

The correct option is B $\frac{21}{40}$

Total balls in bag $I$: $8\to 5$ Red and $3$ Green.

Total balls in bag $II$: $10\to 4$ Red and $6$ Green.

Let $A$ be the event that ball selected from the first bag is red and ball selected from second bag is green.

Let $B$ be the event that ball selected from the first bag is green and ball selected from second bag is red.

$P\left(A\right)=\frac{5}{8}\times \frac{6}{10}=\frac{3}{8}$

$P\left(B\right)=\frac{3}{8}\times \frac{4}{10}=\frac{3}{20}$

Hence, required probability

$P=P\left(A\right)+P\left(B\right)=\frac{3}{8}+\frac{3}{20}=\frac{21}{40}$