Homework SourseAn Xray has a probabilty of 095 of showing a fracture in the
An X-ray has a probabilty of 0.95 of showing a fracture in the leg. If 5 different X-Rays are taken on a particular leg, find the probability that:
a. all five X-rays identify the fracture
b. the fracture does not show up
c. at least 3 X-Rays show the fracture
d. only one X-ray shows the fracture
Solution
a)
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 5
p = the probability of a success = 0.95
x = the number of successes = 5
Thus, the probability is
P ( 5 ) = 0.773780938 [ANSWER]
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b)
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 5
p = the probability of a success = 0.95
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 3.125*10^-7
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c)
Note that P(at least x) = 1 - P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 5
p = the probability of a success = 0.95
x = our critical value of successes = 3
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 2 ) = 0.001158125
Thus, the probability of at least 3 successes is
P(at least 3 ) = 0.998841875 [ANSWER]
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d)
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 5
p = the probability of a success = 0.95
x = the number of successes = 1
Thus, the probability is
P ( 1 ) = 2.96875*10^-5 [ANSWER]
