Question

Box 1 contains 2 red balls and 1 blue ball. Box 2 contains 3 blue balls and 1 red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball.

Answer 1

Answer:

\frac{17}{24}

Step-by-step explanation:

Given that Box 1 contains 2 red balls and 1 blue ball. Box 2 contains 3 blue balls and 1 red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn.

P(selecting Box 1) = 0.5 = P(selecting II box) (assuming a fair coin is tossed)

P(Red/Box I ) = $\frac{2}{2+1} =\frac{2}{3}$

P(Red/Box II) = $\frac{3}{3+1} =\frac{3}{4}$

Box 1 and Box 2 are mutually exclusive and exhaustive events.

So probability of selecting a red ball.

=  probability of selecting a red ball from box 1 +  probability of selecting a red ball frm box 2

$=P(B1)*P(Red/B1) + P(B2)*P(Red/B2)\\= \frac{1}{2} * \frac{2}{3} + \frac{1}{2} * \frac{3}{4} \\= \frac{1}{3} + \frac{3}{8} \\= \frac{17}{24}$