Question

Box 1 contains 3 red 7 blue and Box 2 contains 8 red 2 blue chips. A doe is cast and bowl B1 is selected if 5 or 6 shows up : otherwise bowl B2 is selected. Find the conditional probability of bowl B1 goven that a red chips is drawn

Answer 1

Answer:

The probability that this chip is red is \frac{9}{20} and the conditional probability that the red chip is drawn from bowl II is \frac{2}{3}.

Step-by-step explanation:

(a) There are two ways so this can happen.

First way: Bowl I is selected and then one of the 3 red chips is drawn (from an amount of 10 chips). The probability this happens is given by the product of the probability Bowl I is selected by the probability a red chip is drawn from Bowl I. That is,

\frac{1}{2}\cdot \frac{3}{10} = \frac{3}{20}

Second way: Bowl II is selected and then one of the 6 red chips is drawn (from an amount of 10 chips). The probability this happens is given by the product of the probability Bowl II is selected by the probability a red chip is drawn from Bowl II. That is,

\frac{1}{2}\cdot \frac{6}{10} = \frac{6}{20}

Therefore, the probability that this chip is red is given by the sum of the two probability calculated above, that is, \frac{3}{20}+\frac{6}{20} = \frac{9}{20}.

(b) The probability of A given B is given by

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

Let's A be the event where the selected bowl is bowl II and B be the event where the drawn chip is red.

We have already calculated that P(B) = \frac{9}{20} and, once P(A\cap B) is the probability both event occurs, that is, a red chip is drawn from bowl II, we know that P(A\cap B) = \frac{6}{20}

Therefore, we have P(A|B) = \frac{\frac{6}{20}}{\frac{9}{20}} = \frac{6}{9} = \frac{2}{3}

Step-by-step explanation:

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