Review Day Slide Key

1) Researchers are interested in the relationship between cigarette smoking and lung cancer. Suppose and adult male is randomly selected from a particular population. Assume that the following table shows some probabilities involving the compound event that the individual does or does not smoke and the person is or is not diagnosed with cancer.

EventProbability

Smokes and gets cancer.05

Smokes and does not get cancer.20

Does not smoke and gets cancer.03

Does not smoke and does not get cancer.72

Suppose further that the probability that the randomly selected individual is a smoker is .25.

a) Find the probability that the individual gets cancer, given that he is a smoker.

b) Find the probability that the individual does not get cancer, given that he is a smoker.

c) Find the probability that the individual gets cancer, given that he does not smoke.

d) Find the probability that the individual does not get cancer, given that he does not smoke.

a) P(cancer given smoker) = P(cancer and smoker)/P(smoker) = .05/0.25=.2

b) P(not cancer given smoker) = P(cancer and smoker)/P(smoker) = .20/.25=.8

c) P(cancer given not smoker) = P(cancer and not smoker)/P(not smoker) = .03/.75

d) P(not cancer given not smoker)=P(not cancer and not smoker)/P(not smoker) = .72/.75=.96

- The probability that a car will skid on a bridge on a rainy day is .75. Today the weather station announced that there is a 20% chance of rain. What is the probability that it will rain today and that a car will skid on the bridge.

a) .0300b) .0375c) .1500d) .3000e) .9500

**(.75)(.20) = .1500**

- The probability that Ted will enroll in an English class in 1/3. If he does enroll in an English class, the probability that he would enroll in a math class is 1/5. What is the probability that he enrolls in both classes?

a) 1/15b) 2/15c) 7/15d) 3/5e) 13/15

**4. The local Chamber of Commerce conducted a survey of one thousand randomly selected shoppers at a mall. For all shoppers, “gender of shopper” and “items shopping for” was recorded.**

**Items shopping for**

**GenderClothing ShoesOther Total**

**Male 7525150 250**

**Female 350 230 170 750**

**Total 425 255 320 1000**

**a)What is the probability that the shopper is a female?**

75/1000=.75

**b)What is the probability that the shopper is shopping for shoes?**

255/1000=.255

**c)What is the probability that the shopper is a female shopping for shoes?**

230/1000=.23

**d)What is the probability that the shopper is shopping for shoes given that the shopper is a female?**

Shopping for shoes and female = .23 and female = .75, so .23/.75 = .3067

**e)Are the events “female” and “shopping for shoes” disjoint?**

Shopping for shoes and female = .23 which is not 0 so they are not disjoint.

**f)Are the events “female” and “shopping for shoes” independent?**

Shopping for shoes given female = .3067 and shopping for shoes = .255 which means that those two probabilities are not equal. Therefore “female” and shopping for shoes” are not independent.

5) **Two companies offer overnight shipping that is supposed to arrive by 10:00 a.m. Your company ships 30% of its packages with shipping service 1 and 70% with shipping service 2. Shipping service 1 fails to meet the 10:00 a.m. deadline 10% of the time and shipping service 2 fails to meet it 8% of the time. If a customer is expecting a package and it is late, which of the shipping services is the more likely to have been used? (Hint: use a tree diagram).**

Let S1 = shipping service 1, S2 = shipping service 2. Let S = package arrives by 10am, F = package doesn’t arrive by 10am.

Find P(S1 given F) = P(S1 and F)/P(F) = .03/.086=.349

Find P(S2 given F) = P(S2 and F)/P(F) = .056/.086