Answer to Question #275516 in Statistics and Probability for Subhash

Question

1. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W

bulbs.

i. Suppose that three bulbs are randomly selected. What is the probability that exactly

two of the selected bulbs are rated 75 W?

ii. If two bulbs are randomly selected from the box of lightbulbs and at least one of them

is found to be rated 75 W, what is the probability that both of them are 75-W bulbs?

iii. Given that at least one of the two selected is not rated 75 W, what is the probability

that both selected bulbs have the same rating?

Answer 1

There are, four 40-Watt bulbs, five 60-Watt bulb and six 75-Watt bulbs. The total number of bulbs is 4+5+6=15

a)

To find the probability that out of three randomly selected bulbs there are exactly two 75-Watt bulb is, we apply the combination formula as shown below,

Out of six 75-Watt bulb, we select only two. This is written using combination formula as follows,

$\binom{6}{2}=15$

Out of the remaining 9 bulbs, we only select 1. This is given as,

$\binom{9}{1}=9$

The total number of ways for randomly choosing 3 bulbs from a total of 15 bulbs is,

$\binom{15}{3}=455$

Therefore, the probability exactly two of the 3 selected bulbs are rated 75 W is,

p(exactly two 75-Watt bulb)=(15*9)/455=0.2967 (4dp)

b)

Let the random variable $X$ be the number of 75-Watt bulbs selected.

We find the probability,

$p(X=2|X\geqslant1)=p((X=2)\cap(X\geqslant1))/p(X\geq 1)=p(X=2)/(1-p(X=0))$

Here, $p(X=2)=6/15*5/14=30/210=1/7$ and $p(X=0)=12/210+20/210+20/210+20/210=72/210=36/105$

Therefore,

$p(X=2)/(1-p(X=0))=(1/7)/(1-(36/105))= 0.2173913$

Thus, given that at least one of two selected bulbs is rated 75 W, the probability that both of them are 75-W bulbs is 0.2173913.

c)

Let the random variable $Y$ denote the number of 40 Watt bulb and the random variable $Z$ denote the number of 60-Watt bulb selected. Also, let the random variable $M$ denote the event that both selected bulbs have the same rating. We find the following probability,

$p(M|X\leq1)=p(M\cap X\leq1)/p(X\leq1)=(p(M\cap X=0)+p(M\cap X=1))/(1-p(X=2))$

This probability is also given as,

$(p(M\cap X=0)+0)/(1-p(X=2))=(p(Y=2)+p(Z=2))/(1-p(X=2))=((12/210)+(20/210))/(1-1/7)$

Simplifying this, we have,

(32/210)/(6/7)=8/45

Therefore, Given that at least one of the two selected is not rated 75 W, the probability that both selected bulbs have the same rating is 8/45.