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Bag I contains 3 black and 2 white balls. Bag II contains 2 black and 4 white balls. A bag is selected at random and then a ball is drawn from it. The probability that the ball drawn is black is:

Solution

✅ Correct Option: 1

Bag I contains 3 black and 2 white balls, giving a total of 5 balls.

Bag II contains 2 black and 4 white balls, giving a total of 6 balls.

A bag is selected at random, then a ball is drawn from the selected bag.


The probability of selecting each bag:

P(selecting Bag I) =12= \frac{1}{2}

P(selecting Bag II) =12= \frac{1}{2}


The probability of drawing a black ball from each bag:

P(Black ball from Bag I) =35= \frac{3}{5}

P(Black ball from Bag II) =26= \frac{2}{6}

=13= \frac{1}{3}


A black ball can be drawn in two ways: selecting Bag I and drawing black, or selecting Bag II and drawing black.

P(Black ball) == P(Bag I) ×× P(Black from I) ++ P(Bag II) ×× P(Black from II)

=12×35+12×13= \frac{1}{2} × \frac{3}{5} + \frac{1}{2} × \frac{1}{3}

=310+16= \frac{3}{10} + \frac{1}{6}


The LCM of 10 and 6 is 30.

310=930\frac{3}{10} = \frac{9}{30}

16=530\frac{1}{6} = \frac{5}{30}

930+530=1430\frac{9}{30} + \frac{5}{30} = \frac{14}{30}

=715= \frac{7}{15}


Therefore, the probability that the ball drawn is black is 715\frac{7}{15}.

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