Two urns identical in appearance contain respectively 3 green and 2 black balls and 2 green and 5 black…
Two urns identical in appearance contain respectively 3 green and 2 black balls and 2 green and 5 black balls. One urn is selected at random and a ball is drawn from it. The probability that it is black is
$\frac{39}{70}$
$\frac{37}{70}$
$\frac{41}{70}$
$\frac{33}{70}$
Solution
There are two urns
Urn 1 contains 3 green and 2 black balls.
Urn 2 contains 2 green and 5 black balls.
Let $A$ be event of selecting black ball.
Probability of selecting black ball from urn $1=\frac{2}{5}$
Probability of selecting black ball from urn $2=\frac{5}{7}$
Since, one urn is selected at random hence selecting one urn have probability $\frac{1}{2}$.
$
\therefore P(A)=\frac{1}{2}\left(\frac{2}{5}+\frac{5}{7}\right)=\frac{1}{2}\left(\frac{14+25}{35}\right)=\frac{39}{70}
$