Question

Following are three boxes containing balls. Box #1 contains 2 black balls and 1 white ball. Box #2 contains 2 black balls and 2 white balls. Box #3 contains 1 black ball and 2 white ball. Draw a ball from Box 1 and place it in Box 2. Then draw a ball from Box 2 and place it in Box 3. Finally draw a ball from Box 3 . What is the possibility that the last ball drawn, from Box 3, is black? (Hint: Tree diagram!!)

Answer 1

Let W, B be the event of drawing the white and black balls repectively from Box.

The sample space for last ball drawn, from Box 3, is black are,

(WWB) (WBB) (BWB) (BBB)

P(WWB) = P(Draw white ball from Box #1) * P(Draw white ball from Box #2 | Draw white ball from Box #1) * P(Draw black ball from Box #3 | Draw white ball from Box #2)

= (1/3) * (3/5) * (1/4) = 3/60

P(WBB) = P(Draw white ball from Box #1) * P(Draw black ball from Box #2 | Draw white ball from Box #1) * P(Draw black ball from Box #3 | Draw black ball from Box #2)

= (1/3) * (2/5) * (2/4) = 4/60

P(BWB) = P(Draw black ball from Box #1) * P(Draw white ball from Box #2 | Draw black ball from Box #1) * P(Draw black ball from Box #3 | Draw white ball from Box #2)

= (2/3) * (2/5) * (1/4) = 4/60

P(WBB) = P(Draw black ball from Box #1) * P(Draw black ball from Box #2 | Draw black ball from Box #1) * P(Draw black ball from Box #3 | Draw black ball from Box #2)

= (2/3) * (3/5) * (2/4) = 12/60

Probability that the last ball drawn, from Box 3, is black = 3/60 + 4/60 + 4/60 + 12/60

= 23/60

= 0.3833