Question

Bowl B1 contains two white chips, bowl B2 con-tains two red chips, bowl B3 contains two white and two red chips, and bowl B4 contains three white chips and one red chip. The probabilities of selecting bowl B1, B2, B3, or B4 are 1/2, 1/4, 1/8, and 1/8, respectively. A bowl is selected using these probabilities and a chip is then drawn at random. Find (a) P(W), the probability of drawing a white chip. P(B1 I W), the conditional probability that bowl B1 had been selected, given that a white chip was drawn.

Answer 1

Answer:

By using law of total probability:

P(W) = ∑i P(Bi ) P(W/Bi)

          =1/2 .1 + 1/4 .0 + 1/8 .1/2 + 1/8 .3/4

           =1/2 + 0 + 1/16 + 3/32

            = 21/32

          = 0.656

Step-by-step explanation:

B1 contains = two white chips

B2 contain = 2 red chips

B3 contain = 2 white and 2 red chips

B4 contains = 3 white chips and 1 red chips

P(B1) = 1/2

P(B2) = 1/4

P(B3) = 1/8

P(B4) = 1/8

Probability of drawing white chips:

P(W / B1) = 2. 1/2  = 1

P(W/B2) = 0/2 = 0

P(W/B3) = 2/4 = 2

P(W/B4) =3/4

By using law of total probability:

P(W) = ∑i P(Bi ) P(W/Bi)

          =1/2 .1 + 1/4 .0 + 1/8 .1/2 + 1/8 .3/4

           =1/2 + 0 + 1/16 + 3/32

            = 21/32

          = 0.656