Question

One urn contains 6 blue balls and 14 white balls, and a second urn contains 12 blue balls and 7 white balls. An urn is selected at random, and a ball is chosen from the urn. a. What is the probability that the chosen ball is blue? b. If the chosen ball is blue, what is the probability that it came from the first urn?

Answer 1

Answer:

a) 0.4658 = 46.58% probability that the chosen ball is blue

b) 0.322 = 32.2% probability that it came from the first urn

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

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. What is the probability that the chosen ball is blue?

6/20 = 0.3 of 0.5(first urn)

12/19 = 0.6316 out of 0.5(second urn).

So

$P(A) = 0.3*0.5 + 0.6316*0.5 = 0.4658$

0.4658 = 46.58% probability that the chosen ball is blue.

b. If the chosen ball is blue, what is the probability that it came from the first urn?

Event A: Blue Ball

Event B: From first urn

From item a., $P(A) = 0.4658$

Probability of blue ball from first urn:

0.3 of 0.5. So

$P(A \cap B) = 0.3*0.5 = 0.15$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.4658} = 0.322$

0.322 = 32.2% probability that it came from the first urn