8.4
The probability that a male will be color-blind is 0.53. Find the probabilities that in a group of 39 men, the following will be true. (a) Exactly 5 are color-blind. (b) No more than 5 are color-blind. (c) At least 1 is color-blind.
a) n=39, x=5, p=0.053. P(5)=C(39,5)•(0.053)5•(1−0.053)34 ≈0.0378 b) P(no more than 5)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5) =C(39,0)•(0.53)^0•(1−0.53)^39−0 +C(39,1)•(0.53)^1•(1−0.53)^39−1 +C(39,2)•(0.53)^2•(1−0.53)^39−2 +C(39,3)•(0.53)^3•(1−0.53)^39−3 +C(39,4)•(0.53)^4•(1−0.53)^39−4 +C(39,5)•(0.53)^5•(1−0.53)^39−5 P(no more than 5)≈0.9840 c) P(at least 1) = 1−P(0) = 1−C(39,0)•(0.053)^0•(1−0.053)^39 ≈0.8804
According to an airline, a particular flight is on time 92% of the time. Suppose 52 flights are randomly selected and the number of on time flights is recorded. Find the probabilities of the following events occurring. a. All 52 flights are on time b. Between 46 and 48 flights (inclusive) are on time
a) C(52,52)•(0.92)^52•(1−0.92)^0≈0.0131 b) The probability of any of these events happening is the sum of the individual probabilities. Simplify this sum, rounding to four decimal places. C(52,46)•(0.92)^46•(1−0.92)^52−46+C(52,47)•(0.92)^47•(1−0.92)^52−47 +C(52,48)•(0.92)^48•(1−0.92)^52−48≈0.4870
93% of Dr.William's patients end up with20-30 vision or better. Find the probability that exactly 5 of her next 6 patients end up with20-30 vision or better.
answer= .2922 Consider a 'success' to be a patient ending with 20-30 vision or better. n=6, x=5, p=0.93 P(x successes in n trials)=C(n,x)•px•(1−p)n−x = C(6,5)•(.93)^5•(1−(.93))^6-5
A survey finds that customers are charged incorrectly for 1 out of every 30 items, on average. Suppose a customer purchases 13 items. Find the following probability. A customer is not charged incorrectly for any item.
answer= .6436 P(x successes in n trials)=C(n,x)•px•(1−p)n−x = C(13,0)•(1/30)^0•(1−(1/30))^13−0 n = 13 In this case, n is the number of items purchased = 13 Since customers are charged incorrectly for 1 out of every 30 items, the probability that a customer is charged incorrectly, p, is 1/30.
According to a survey, 22.4% of credit-card-holding families in a certain area hardly ever pay off the balance. Suppose a random sample of 20 credit-card-holding families is taken. Find the probability that at least 4 families hardly ever pay off the balance. The probability that at least 4 families hardly ever pay off the balance is?
p = .224 failure = 1-.224 n=20 x=4 Use the binomial probability formula to calculate the probabilites that 0, 1, 2, and 3 families hardly ever pay off the balance. Find P(0) first, rounding to four decimal places. Then add all together to get and subtract from 1 to get answer 1 - ( P(0) + P(1) + P(2) + P(3) ) answer = .6864
A survey finds customers are charged incorrectly for 2 out of every 10 items. Suppose a customer purchases 12 items. Find the probability that the customer is charged incorrectly on at most 2 items.
p = 2/10 = .2 n = 12 The probability that the customer is charged incorrectly on at least 3 items is given by the following: 1−P(is charged incorrectly on 0, 1, or 2items)
A recent study found that 87% of breast-cancer cases are detectable by mammogram. Suppose a random sample of 17 women with breast cancer are given mammograms. Find the probability that all of the cases are detectable, assuming that detection in the cases is independent. Substitute n=17, x=17, and p=0.87 into the equation.
.P(17 successes in 17 trials) =C(17,17)•0.8717•(1−0.87)17−17 ≈ 0.0937
A factory tests a random sample of 25 transistors for defects. The probability that a particular transistor will be defective has been established by past experience as 0.06. p = 1-.06 = .94 n = 31 x = 31 What is the probability that there are no defective transistors in the sample?
C(25,25)(.94)^25(1-.94)^25-25 = .2129
