Question

Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank extracts 2 balls with replacement. USE ONLY THE CONDITIONAL PROBABILITY FORMALISM and derive an expression for the probability that the 2 extracted balls are red.

Answer 1

Complete question:

Consider 3 boxes, each of which contains 4 balls in particular, box 1 contains 4 white balls, box 2 contains 3 white balls and 1 red ball, box 3 contains 2 white balls and 2 red balls. Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank extracts 2 balls with replacement. USE ONLY THE CONDITIONAL PROBABILITY FORMALISM and derive an expression for the probability that the 2 extracted balls are red.

Answer:

11/288

Step-by-step explanation:

We are given:

Box 1: ( 4White, ORed)

Box 2: (3White, 1Red)

Box 3: (2White, 2 Red)

We are told that if Frank choose from Box 1 and Box 3, 2 balls are extracted without replacement.

Since there is no red ball in Box 1, there is no way 2 red balls will come out from Box1.

Our Event, E = getting 2 red balls.

Now Box 1 is ruled out, we have:

P[E(B1)]= 0

P[E/B3)] = (2 2) / (4 2)

= 1/6

If box 2 is chosen, 2 balls are extracted with replacement. Therefore for Box 2:

P(E/B2) = (1/4) *(1/4)

= 1/16

Therefore probability that 2 balls extracted are red, we have:

P(E)=P(E/B1) P(B1) + P(E/B2) P(B2)+P(E/B3) P(B3)

$= 0 * \frac{1}{3} + \frac{1}{6}*\frac{1}{3} + \frac{1}{16}*\frac{1}{3}$

= 11/288