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

\frac{17}{24}

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 ) =
`(2)/(2+1) =(2)/(3)`

P(Red/Box II) =
`(3)/(3+1) =(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)\\= (1)/(2) * (2)/(3) + (1)/(2) * (3)/(4) \\= (1)/(3) + (3)/(8) \\= (17)/(24)`