Answer to Question #96162 in Statistics and Probability for fatima

Question

Two boxes each contain red and white colored bulbs. The percentage of red bulbs in the first box is 30% and 10% in the second box. One box is chosen at random and a bulb is withdrawn.

a. Construct a probability tree and show all relevant probabilities.
b. What is the probability of withdrawing a red bulb?
c. If a red bulb is withdrawn, what is the probability that it came from box 2?

Answer 1

Let $I$ be the event that the first box is chosen, $II$ be the event that the second box is chosen, $R$ be the event that the red bulb is chosen, and $W$ be the event that the white bulb is chosen.

Given that

$$P(I)=P(II)=0.5, P(R\ | \ I)=0.3, P(R\ |\ II)=0.1$$

Construct a probability tree and show all relevant probabilities. 

We have that

$$P(W\ | \ I)=1-P(R\ | \ I)=0.7,$$ $$P(W\ | \ II)=1-P(R\ |\ II)=0.9$$

Then

$$P(I\cap R)=P(I)P(R\ | \ I)=0.5(0.3)=0.15$$ $$P(I\cap W)=P(I)P(W\ | \ I)=0.5(0.7)=0.35$$ $$P(II\cap R)=P(II)P(R\ | \ II)=0.5(0.1)=0.05$$ $$P(II\cap W)=P(II)P(W\ | \ II)=0.5(0.9)=0.45$$

b. What is the probability of withdrawing a red bulb?

By the Theorem of total probability

$$P(R)=P(I)P(R\ | I)+P(II)P(R\ | \ II)$$

$$P(R)=0.5(0.3)+0.5(0.1)=0.2$$

c. If a red bulb is withdrawn, what is the probability that it came from box 2?

By the Bayes' rule

$$P(II \ | \ R)={P(II)P(R\ | \ II) \over P(I)P(R\ | I)+P(II)P(R\ | \ II)}$$

$$P(II \ | \ R)={0.5(0.1) \over 0.5(0.3)+0.5(0.1)}=0.25$$