Math 2030 2030 M M Homework Homework 22 Math Answers and and Solutions Solutions Answers 4. Name Name the the events events A A and and B. B. PP(A) (A) = = .1 .1 and and PP(B) (B) = = .3. .3. As As A A and and B B are are independent, independent, so so are are A A and and B Bcc,, 4. c c c c c c A and and B, B, A A and and B B .. Can Can you you prove prove this? this? A (a) (a) PP(A (AccB Bcc)) = = (.9)(.7) (.9)(.7) = = .63 .63.. cc cc (b) 1 − P (A B ) = .37 . (b) 1 − P (A B ) = .37 . c c (c) (c) PP(AB (AB c)) + + PP(A (AcB) B) = = (.1)(.7) (.1)(.7) + + (.9)(.3) (.9)(.3) = = .34 .34.. 5. 5. (a) (a) and and (b) (b)

choose urn 1

2/5

3/5

1/2

4/7

draw black

1/5

draw white

3/10

draw black

2/7

draw white

3/14

1/2

choose urn 2

6. 6. 7. 7.

3/7

(c) P (draw black) = 1/5 + 2/7 = 17/35 . (c) P (draw black) = 1/5 + 2/7 = 17/35 . 13·25 P ( 2nd spade AND 1st black ) 25 13·25 52·51 P ( 2nd spade | 1st black ) = P ( 2nd spade AND 1st black ) = 52·51 = 25 . P ( 2nd spade | 1st black ) = = 112 = 102 . P ( 1st black ) P ( 1st black ) 102 2 We have P (A) = .5 and P (A ∪ B) = .8 . We have P (A) = .5 and P (A ∪ B) = .8 . (a) IfIf A A and and B B are are mutually mutually exclusive, exclusive, PP(AB) (AB) = = 0. 0. (a) As P (A ∪ B) = P (A) + P (B) − P (AB), .8 = .5 + + PP(B) (B) − − 0, 0, giving giving PP(B) (B) = = .3 .3.. As P (A ∪ B) = P (A) + P (B) − P (AB), .8 = .5

(b) For For independence, independence, PP(AB) (AB) = = PP(A)P (A)P(B). (B). Then Then (b) P (A ∪ B) = P (A) + P (B) − P (A)P (B) gives .8 = = .5 .5 + + PP(B) (B) − − (.5)P (.5)P(B) (B) from from which which PP(B) (B) = = .6 .6 P (A ∪ B) = P (A) + P (B) − P (A)P (B) gives .8 11

8. Here is a solution using the formulas. P (card BW | see B ) =

P ( card BW AND see B ) = P ( see B )

P ( card BW AND see B ) = P ( see B AND card WW ) + P ( see B AND card BW ) + P ( see B AND card BB ) P ( see B | card BW )P ( card BW ) P ( see B | card WW )P (card WW) + P ( see B | card BW )P (card BW) + P ( see B | card BB )P (card BB) =

(.5)(.5) 5 = (0)(.3) + (.5)(.5) + (1)(.2) 9

Alternatively, consider a hat with 10 cards, 3 WW, 5 BW, 2 BB. Reach into the hat, take a card at random, and place it on the table. What is visible to you is one 20 possible card sides, any one as likely to be seen as any other. 9 are black. Note that 5 of those 9 have white on their reverse side. The probability that a card is BW given that you see B is 5/9. The proposed “solution” is wrong. It construes the condition as knowing that at least one side of the card you chose is black. Had you reached into the hat and before looking at anything were told that ”at least one side of the card in your hand is black” you would be correct in calculating the probability that your card has a white side as 5/7. This is subtle. It is not enough to know the content of the condition (in this case, construed as there being a black side) but also the means by which that content is obtained. Well written textbook style questions make these means clear. The Bar-Hillel, Falk article investigates what can happen when a problem which appears clear in ordinary language leaves those means ambiguous. 3. Let R be the event ”it rains”, F the event ”rain is forecast”, U be the event, ”Mr. Pickwick takes his ¯ = 1/2, P (R|F) = 2/3, P (R|F) ¯ ¯ = 1/3, umbrella”. We are given, P (R) = 1/2, P (R) = 1/3, P (R|F) ¯ F) ¯ = 2/3, P (U|F) = 1, P (U|F) ¯ ¯ = 1/3, P (U| ¯ F) ¯ = 2/3. P (R| = 0, P (U|F) ¯ ¯ Part (a) asks for P (U|R) and (b) for P (U|R). Begin by determining P (F). From P (F|R) = P (R|F)P (F)/P (R), P (F|R) = (4/3)P (F). ¯ ¯ (F)/P ¯ ¯ ¯ Similarly, P (F|R) = P (R|F)P (R), giving P (F|R) = (2/3)P (F). ¯ Then as P (F|R) + P (F|R) = 1, (4/3)P (F) + (2/3)(1 − P (F)) = 1 and P (F) = 1/2. ¯ = 1/2, and P (F|R) = 2/3 and P (F|R) ¯ It follows that P (F) = 1/3. ¯ = 1/3 and P (F| ¯ R) ¯ = 2/3. Similarly, P (F|R) ¯ ¯ ∩ R)/P (R) = 2 P (U ¯ ∩ R). (a) P (U|R) = P (U ¯ ∩ R) = P (U ¯ ∩F ¯ ∩ R) + P (U ¯ ∩ F ∩ R) = P (U| ¯ F ¯ ∩ R)P (F|R)P ¯ ¯ ∩ R)P (F|R)P (R) P (U (R) + P (U|F ¯ =(2/3) · (1/3) · (1/2) + 0 · (2/3) · (1/2) = 1/9 and P (U|R) = 2/9. ¯ = P (U ∩ R)/P ¯ ¯ = 2 P (U ∩ R). ¯ (b) P (U|R) (R) ¯ ¯ ¯ ¯ = P (U|F ∩ R)P ¯ (F|R)P ¯ (R) ¯ + P (U|F ¯ ∩ R)P ¯ (F| ¯ R)P ¯ (R) ¯ P (U ∩ R) = P (U ∩ F ∩ R) + P (U ∩ F ∩ R) ¯ =1 · (1/3) · (1/2) + (1/3) · (2/3) · (1/2) = 5/18 and P (U|R) = 5/9.

2

It is also possible to solve this problem using counting. Over 72 days (chosen at random), let 3f be the number of days rain is forecast. Then 2f + (1/3)(72 − 3f ) = 36, the expected number of days of rain. Solve to get f = 12 so rain would be forecast on 36 days, half the time. This gives the following table. Forecast Rain Forecast No Rain

Rain 24 12

No Rain 12 24

Incorporate the number of days on which an umbrella is taken. Forecast Rain Forecast No Rain

Rain 24 (take umbrella on 24) 12 (take umbrella on 4)

Finally, solve by evaluating the appropriate ratios. ¯ For (a), P (U|R) = 1 − P (U|R) = 1 − (28/36) = 2/9. ¯ = 20/36 = 5/9. For (b), P (U|R)

3

No Rain 12 (take umbrella on 12) 24 (take umbrella on 8)

choose urn 1

2/5

3/5

1/2

4/7

draw black

1/5

draw white

3/10

draw black

2/7

draw white

3/14

1/2

choose urn 2

6. 6. 7. 7.

3/7

(c) P (draw black) = 1/5 + 2/7 = 17/35 . (c) P (draw black) = 1/5 + 2/7 = 17/35 . 13·25 P ( 2nd spade AND 1st black ) 25 13·25 52·51 P ( 2nd spade | 1st black ) = P ( 2nd spade AND 1st black ) = 52·51 = 25 . P ( 2nd spade | 1st black ) = = 112 = 102 . P ( 1st black ) P ( 1st black ) 102 2 We have P (A) = .5 and P (A ∪ B) = .8 . We have P (A) = .5 and P (A ∪ B) = .8 . (a) IfIf A A and and B B are are mutually mutually exclusive, exclusive, PP(AB) (AB) = = 0. 0. (a) As P (A ∪ B) = P (A) + P (B) − P (AB), .8 = .5 + + PP(B) (B) − − 0, 0, giving giving PP(B) (B) = = .3 .3.. As P (A ∪ B) = P (A) + P (B) − P (AB), .8 = .5

(b) For For independence, independence, PP(AB) (AB) = = PP(A)P (A)P(B). (B). Then Then (b) P (A ∪ B) = P (A) + P (B) − P (A)P (B) gives .8 = = .5 .5 + + PP(B) (B) − − (.5)P (.5)P(B) (B) from from which which PP(B) (B) = = .6 .6 P (A ∪ B) = P (A) + P (B) − P (A)P (B) gives .8 11

8. Here is a solution using the formulas. P (card BW | see B ) =

P ( card BW AND see B ) = P ( see B )

P ( card BW AND see B ) = P ( see B AND card WW ) + P ( see B AND card BW ) + P ( see B AND card BB ) P ( see B | card BW )P ( card BW ) P ( see B | card WW )P (card WW) + P ( see B | card BW )P (card BW) + P ( see B | card BB )P (card BB) =

(.5)(.5) 5 = (0)(.3) + (.5)(.5) + (1)(.2) 9

Alternatively, consider a hat with 10 cards, 3 WW, 5 BW, 2 BB. Reach into the hat, take a card at random, and place it on the table. What is visible to you is one 20 possible card sides, any one as likely to be seen as any other. 9 are black. Note that 5 of those 9 have white on their reverse side. The probability that a card is BW given that you see B is 5/9. The proposed “solution” is wrong. It construes the condition as knowing that at least one side of the card you chose is black. Had you reached into the hat and before looking at anything were told that ”at least one side of the card in your hand is black” you would be correct in calculating the probability that your card has a white side as 5/7. This is subtle. It is not enough to know the content of the condition (in this case, construed as there being a black side) but also the means by which that content is obtained. Well written textbook style questions make these means clear. The Bar-Hillel, Falk article investigates what can happen when a problem which appears clear in ordinary language leaves those means ambiguous. 3. Let R be the event ”it rains”, F the event ”rain is forecast”, U be the event, ”Mr. Pickwick takes his ¯ = 1/2, P (R|F) = 2/3, P (R|F) ¯ ¯ = 1/3, umbrella”. We are given, P (R) = 1/2, P (R) = 1/3, P (R|F) ¯ F) ¯ = 2/3, P (U|F) = 1, P (U|F) ¯ ¯ = 1/3, P (U| ¯ F) ¯ = 2/3. P (R| = 0, P (U|F) ¯ ¯ Part (a) asks for P (U|R) and (b) for P (U|R). Begin by determining P (F). From P (F|R) = P (R|F)P (F)/P (R), P (F|R) = (4/3)P (F). ¯ ¯ (F)/P ¯ ¯ ¯ Similarly, P (F|R) = P (R|F)P (R), giving P (F|R) = (2/3)P (F). ¯ Then as P (F|R) + P (F|R) = 1, (4/3)P (F) + (2/3)(1 − P (F)) = 1 and P (F) = 1/2. ¯ = 1/2, and P (F|R) = 2/3 and P (F|R) ¯ It follows that P (F) = 1/3. ¯ = 1/3 and P (F| ¯ R) ¯ = 2/3. Similarly, P (F|R) ¯ ¯ ∩ R)/P (R) = 2 P (U ¯ ∩ R). (a) P (U|R) = P (U ¯ ∩ R) = P (U ¯ ∩F ¯ ∩ R) + P (U ¯ ∩ F ∩ R) = P (U| ¯ F ¯ ∩ R)P (F|R)P ¯ ¯ ∩ R)P (F|R)P (R) P (U (R) + P (U|F ¯ =(2/3) · (1/3) · (1/2) + 0 · (2/3) · (1/2) = 1/9 and P (U|R) = 2/9. ¯ = P (U ∩ R)/P ¯ ¯ = 2 P (U ∩ R). ¯ (b) P (U|R) (R) ¯ ¯ ¯ ¯ = P (U|F ∩ R)P ¯ (F|R)P ¯ (R) ¯ + P (U|F ¯ ∩ R)P ¯ (F| ¯ R)P ¯ (R) ¯ P (U ∩ R) = P (U ∩ F ∩ R) + P (U ∩ F ∩ R) ¯ =1 · (1/3) · (1/2) + (1/3) · (2/3) · (1/2) = 5/18 and P (U|R) = 5/9.

2

It is also possible to solve this problem using counting. Over 72 days (chosen at random), let 3f be the number of days rain is forecast. Then 2f + (1/3)(72 − 3f ) = 36, the expected number of days of rain. Solve to get f = 12 so rain would be forecast on 36 days, half the time. This gives the following table. Forecast Rain Forecast No Rain

Rain 24 12

No Rain 12 24

Incorporate the number of days on which an umbrella is taken. Forecast Rain Forecast No Rain

Rain 24 (take umbrella on 24) 12 (take umbrella on 4)

Finally, solve by evaluating the appropriate ratios. ¯ For (a), P (U|R) = 1 − P (U|R) = 1 − (28/36) = 2/9. ¯ = 20/36 = 5/9. For (b), P (U|R)

3

No Rain 12 (take umbrella on 12) 24 (take umbrella on 8)