A box in a certain supply room contains five 40-W lightbulbs, four 60-W bulbs, a
ID: 3132165 • Letter: A
Question
A box in a certain supply room contains five 40-W lightbulbs, four 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected. (Round your answers to four decimal places.) What is the probability that exactly two of the selected bulbs are rated 75-W? What is the probability that all three of the selected bulbs have the same rating? What is the probability that one bulb of each type is selected? Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?Explanation / Answer
A)
There are 15 bulbs, of which 6 are 75W.
Note that the probability of x successes out of n trials is
P(x) = C(N-K, n-x) C(K, x) / C(N, n)
where
N = population size = 15
K = number of successes in the population = 6
n = sample size = 3
x = number of successes in the sample = 2
Thus,
P( 2 ) = 0.296703297 [ANSWER]
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b)
FOR ALL 40W:
Note that the probability of x successes out of n trials is
P(x) = C(N-K, n-x) C(K, x) / C(N, n)
where
N = population size = 15
K = number of successes in the population = 5
n = sample size = 3
x = number of successes in the sample = 3
Thus,
P( 3 ) = 0.021978022
FOR ALL 60W:
Note that the probability of x successes out of n trials is
P(x) = C(N-K, n-x) C(K, x) / C(N, n)
where
N = population size = 15
K = number of successes in the population = 4
n = sample size = 3
x = number of successes in the sample = 3
Thus,
P( 3 ) = 0.008791209
FOR ALL 75W:
Note that the probability of x successes out of n trials is
P(x) = C(N-K, n-x) C(K, x) / C(N, n)
where
N = population size = 15
K = number of successes in the population = 6
n = sample size = 3
x = number of successes in the sample = 3
Thus,
P( 3 ) = 0.043956044
Hence,
P(all same) = 0.021978022+0.008791209+0.043956044 = 0.074725275 [ANSWER]
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c)
There are 5*4*6 = 120 ways to do that.
There are 15C3 = 455 ways to get any 3 bulbs.
Hence,
P = 120/455 = 0.263736264 [ANSWER]
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d)
This is the probability that the first 5 bulbs are not 75W.
Note that the probability of x successes out of n trials is
P(x) = C(N-K, n-x) C(K, x) / C(N, n)
where
N = population size = 15
K = number of successes in the population = 6
n = sample size = 5
x = number of successes in the sample = 0
Thus,
P( 0 ) = 0.041958042 [ANSWER]
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