Question

Bowl B1 contains one white and two red chips; bowl B2 contains two white and two red chips; bowl B3 contains one white and four red chips. The probabilities of selecting bowl B1, B2, or B3 are 0.3, 0.2, and 0.5, 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.
b. P(B1 Given W), the conditional probability that bowl B1 had been selected, given that a white chip was drawn.

Answer 1

Answer:

a) 0.3 = 30% probability of drawing a white chip.

b) 0.3333 = 33.33% probability that bowl B1 had been selected, given that a white chip was drawn.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

a. P(W), the probability of drawing a white chip.

1/3(one white out of 3) of 0.3(from B1).

1/2(two white out of 4) of 0.2(from B2).

1/5(one white out of 5) of 0.5(from B3). So

$P(W) = 0.3333*0.3 + 0.5*0.2 + 0.2*0.5 = 0.3$

0.3 = 30% probability of drawing a white chip.

b. P(B1 Given W), the conditional probability that bowl B1 had been selected, given that a white chip was drawn.

The probability of drawing a white chip from B1 is 1/3 out of 0.3, so:

$P(B1 \cap W) = 0.3\frac{1}{3} = 0.1$

Then the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.3} = 0.3333$

0.3333 = 33.33% probability that bowl B1 had been selected, given that a white chip was drawn.