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1. 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. $\frac{19}{20}$

B. $\frac{17}{20}$

C. $\frac{8}{10}$

D. $\frac{21}{40}$

Solution:

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.

\begin{aligned} P (A) = (\frac{5}{8}) \times (\frac{6}{10}) \\ = \frac{3}{8} and \\ P (B) = (\frac{3}{8}) \times (\frac{4}{10}) \\ = \frac{3}{20} \end{aligned}

Hence, required probability,

\begin{aligned} = P (A) + P (B) \\ = \frac{3}{8} + \frac{3}{20} \\ = \frac{21}{40} \end{aligned}

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