PROBLEM 2 (20 points):

The Gift Basket Store had the following premade gift baskets containing the following combinations in stock:

Cookies Mugs Candy

Coffee 20 13 10

Tea 12 10 12

Choose one basket at random.Find the probability that it contains

a)Tea or Cookies

b)Coffee given that it contains Candies

c)Tea and Mugs

d)Is there any relation between picking cookies and the type of drink (coffee, tea)?

(consider P(cookies|coffee), P(cookies|tea), P(cookies))

example of solution:

Justify your answers. Show the literal formula and not just the final result.

PROBLEM

A survey examines customers' preferences in having a sunroof in their car.

Sunroof (S)

No Sunroof (S')

Woman (W)

800

650

1450

Man (M)

1500

300

1800

2300

950

3250

a) Compute the Marginal Probabilities and the Joint Probabilities.

b) Compute: P(W), P(S|M), P(S'|W), P(M and S), P(W or S).

c) Is there any relationship between sex (woman or man) and preferring to have a sunroof?

SOLUTION

a) Compute the Marginal Probabilities and the Joint Probabilities.

Marginal

Joint

P(S) = 2300/3250 = 71%

P(S and W) = 800/3250 = 25%

P(S') = 950/3250 = 29%

P(S and M) = 1500/3250 = 46%

P(W) = 1450/3250 = 45%

P(S' and W) = 650/3250 = 20%

P(M) = 1800/3250= 55%

P(S' and M) = 300/3250 = 9%

b) Compute: P(W), P(S|M), P(S'|W), P(M and S), P(W or S) .

P(W) =45%

P(S|M) = P(S and M)/P(M) = 46%/55% = 84%

P(S'|W) = P(S' and W)|P(W)=44%

P(M and S) =46%

P(W or S) = P(W)+P(S)-P(W and S) = 45% + 71% - 25% = 91%

c) Is there any relationship between sex (woman or man) and wanting to have a sunroof?

P(S) = ? P(S | W) =? P(S|M)

P(S) = 71%P(S|W) = 55% P(S|M) = 84%

If we don't have equality among the above probabilities, then there IS a relationship.

If all the above probabilities would have been equal, then it would have been no relationship.

In Conclusion: There is a greater probability of owning a car with sunroof given you are a man.

Note: discussing relationship should be done based on the given probabilities values (A|B) and not just yes or no.

Another way of testing:

Multiplication rule.If two events ARE INDEPENDENT then:

P(A and B) = P(A)*P(B) // P(A and B) = P(A|B) P(B) = P(B|A) P(A)

Note: If you have an intersection between the events do not use the above rule.

Addition Rule. The correct way to compute P(A or B):

P(A or B) = P(A) + P(B) - P(A and B)