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 Conditional Probability Assignment

 Example 1: Weather records show that on a given type of day, there is a 30% chance of rain, a

40% chance of it being windy, and a 46% chance that it either rains or is windy.

Exercise 1: If you get up on a day as in Example 1, and find that it is raining, what would be

your new estimate for the chance of it being windy during the day?

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 Exercises 2-4 refer to the experiment of rolling a fair die twice. All of the 36 outcomes in the

sample space S are equally likely. E : The first face is odd. F : The sum of the faces is greaterthan 8. G: The sum of the faces is even.

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Exercise 2: Compute P(E|F) and P(F|E).

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Exercise 3: Compute P(E|G) and P(G|E).

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Exercise 4: Compute P(F|G) and P(G|F).

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 Exercises 5-7 refer to the experiment of rolling a fair die twice. All of the 36 possible outcomesare equally likely.  E : The first face is odd. F : The sum of the faces is greater than 8. G: The

sum of the faces is even. Your work on Exercises 2-4 of the section Definitions will be helpful

for these exercises. Refer to the sample space, S, given before Exercise 2.

Exercise 5: Are E and F independent? No, E and F are dependent.

Method 1:  

 

Exercise 6: Are E and G independent? Yes, E and G are independent.

Method 1:      

Exercise 7: Are F and G independent? No, F and G are dependent. 

Method 1:      

Exercise 8: You are going to pick a random sample of three parts from a large box of parts, 5%

of which are defective. After each draw you note whether or not the selected part is defective

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 and then replace it. Hence, there is a 5% chance of getting a defective part on each of the three

draws and the results of the draws are independent of each other. (i) What is the probability thatyou get three non-defective parts? (ii) Explain how you used independence in your answer to

Part i.

(i)P ( D1 ) = .05 = Getting a defective part on 1st

draw

P ( D1C ) = 1- P(D1) = .95 = Not getting a defective part on 1st draw

P ( D2 ) = .05 = Getting a defective part on 2nd draw

P ( D2C ) = 1- P(D2) = .95 = Not getting a defective part on 2nd draw

P ( D3 ) = .05 = Getting a defective part on 3rd draw

P ( D3C

) = 1- P(D3) = .95 = Not getting a defective part on 3rd

draw

P(D1C D2

C D3C

) = P(.95)* P(.95)*P(.95) = .86

(ii) Each of the outcomes was independent of each other since the defective part was put back into the box. 5% was the chance of getting a defective part therefore 95% was the chance of 

getting a working part. (.05 +.95=1) Since the events were independent then I multiplied .95three times to get the answer.