Question

Bowl I contains 3 red chips and 7.bluechips.Bowl II contains 6 red chips
and 4 blue chips.A bowl is selected at random and then I chip is drawn
from this bow1.
(a) Compute the probability that this chip is red.
(b) Relative to the hypothesis that the chip is red, find the conditional
probability that it is drawn from bowl II.

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 $P(A|B) = \frac{\frac{6}{20}}{\frac{9}{20}} = \frac{6}{9} = \frac{2}{3}$.