Question

A box in a certain supply room contains four 30W lightbulbs, five 60W lightbulbs, and six 70W lightbulbs. Suppose that four lightbulbs are randomly selected. 1) What is the probability that at least one lightbulb of each type is selected? 2)Find the expected number of lightbulbs (incorrect ones) that need to be selected before a 60W or 70W lightbulb is selected. Show this by building a probability table.

Answer 1

The 3 types of bulbs are represented here as:
30W lightbulbs: A
60W lightbulbs: B
70W lightbulbs: C

1) The probability that at least one lightbulb of each type is selected is computed here as:
= P(AABC) + P(ABBC) + P(ABCC)

Therefore 0.5275 is the required probability here.

2) There are four bulbs which are not a 60W or a 70W bulb and 11 which are a 60W o a 70W bulb. We stop when we obtain a 60W or a 70W bulb.

Let X be the event that a 60 or a 70 W bulb is drawn and Y be the event that a 30W bulb is drawn. Then the various combinations here are obtained as:

• X, Prob. = 11/15
• YX, Prob = (4/15)*(11/14)
• YYX, Prob. = (4/15)*(3/14)*(11/13)
• YYYX, Prob. = (4/15)*(3/14)*(2/13)*(11/12)
• YYYYX, Prob. = (4/15)*(3/14)*(2/13)*(1/12)

These are the only possibilities here.

The expected value of the incorrect ones withdrawn is computed here as:

E(Y) = 0*(11/15) + 1*(4/15)*(11/14) + 2*(4/15)*(3/14)*(11/13) + 3* (4/15)*(3/14)*(2/13)*(11/12) + 4*(4/15)*(3/14)*(2/13)*(1/12)

= 1/3

Therefore 1/3 is the expected number of incorrect ones here.