1. ## Need Help

If anyone can take a look at these two problems and see my answers and let me know if I did it correct or am doing it wrong it would be greatly appreciated.

1. A certain sports car comes equipped with either an automatic or a manual transmission, and the car is available in one of four colors. Relevant probabilities for various combinations of transmission type and color are given in the accompanying table.

Color
Transmission Type White Blue Black Red
A .13 .10 .11 .11
M .15 .07 .15 .18

Let A = (automatic transmission), B = {black}, and C = {white}

a) Calculate P(A), P(B), and P(A B).
b) Calculate both P(A|B) and P(B|A), and explain in context what each of these probabilities represents

2. Suppose we toss a fair die twice. Let E denote the event that the sum of the two tosses is 8, and F denote the event that the first toss equals 3. Are these two events, E and F, independent? Please use the definition of independence to answer this question.

My solutions:

2. ## Re: Need Help

I believe 1 is correct. As far as the explain in context part, they are probably looking for "42% of the black cars are automatic."

3. ## The Following User Says Thank You to asterisk For This Useful Post:

bigbob (09-29-2013)

4. ## Re: Need Help

2) When you roll a die twice (or two dice once), the sample space is 36

{11, 12, 13, 14, 15, 16
21, 22, 23, 24, 25, 26
31, 32, 33, 34, 35, 36
41, 42, 43, 44, 45,46
51, 52, 53, 54, 55, 56
61, 62, 63, 64, 65, 66}

So P(F) = 5/36

A formal definition of independence is:

Events E and F are independent if P(F|E) = P(F)

Calculate P(F|E)

5. ## The Following User Says Thank You to asterisk For This Useful Post:

bigbob (09-29-2013)

6. ## Re: Need Help

Originally Posted by asterisk
2) When you roll a die twice (or two dice once), the sample space is 36

{11, 12, 13, 14, 15, 16
21, 22, 23, 24, 25, 26
31, 32, 33, 34, 35, 36
41, 42, 43, 44, 45,46
51, 52, 53, 54, 55, 56
61, 62, 63, 64, 65, 66}

So P(F) = 5/36

A formal definition of independence is:

Events E and F are independent if P(F|E) = P(F)

Calculate P(F|E)
Had I a feeling I did the second problem incorrect. Using the P(E)= 5/36 (.138) and P(F)= .166 we have P(E|F)= P(E&F)/P(F), thus P(E|F)= (.138)*(.166)/(.166)= .138 So events E and F are in fact independent.

7. ## Re: Need Help

Originally Posted by bigbob
Had I a feeling I did the second problem incorrect. Using the P(E)= 5/36 (.138) and P(F)= .166 we have P(E|F)= P(E&F)/P(F), thus P(E|F)= (.138)*(.166)/(.166)= .138 So events E and F are in fact independent.
But I cant assume independence between E and F to just multiply them together. How can I find P(E&F) or P(E or F)?

8. ## Re: Need Help

I did if F occurs than P(E)= 1/6 (since we need a 5 rolled now) thus P(E&F)=1/6*1/6=1/36= .027. Is this wrong or correct?

9. ## Re: Need Help

"if F occurs than P(E)= 1/6" is another way of saying P(F|E) = 1/6

In English, P(F|E) is the probability of rolling an 8 given that the first roll is a 3, which you correctly figured out is 1/6

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