A box containing 4 white pens and 2 black pens. Another box containing 3 white pens and 5 black pens

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14.Probability

easy

A box containing $4$ white pens and $2$ black pens. Another box containing $3$ white pens and $5$ black pens. If one pen is selected from each box, then the probability that both the pens are white is equal to

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{1}{5}$

Solution

(c) Total number of pens in first bag = $4 + 2 = 6$
and total number of pens in second bag = $3 + 5 = 8$.
The probability of selecting a white pen from first bag = $\frac{4}{6} = \frac{2}{3}$ and probability of selecting a white pen from second bag = $\frac{3}{8}$.
$\therefore $ Required probability that both the pens are white =$\frac{2}{3} \times \frac{3}{8} = \frac{1}{4}$.

Standard 11

Mathematics