Mathematics 426 Robert Gross Homework 3 Answers P(AB) = P(A) + P(B) P(A B) = 0.44 P(AC) = P(A) + P(C) P(A C) = 0.45 P(BC) = P(B) + P(C) P(B C) = 0

Mathematics 426 Robert Gross Homework 3 Answers 1. Suppose that A, B, and C are events, and you know that P(A) = 0.72 P(A ∪ B) = 0.78 P(B) = 0.50 P...
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Mathematics 426 Robert Gross Homework 3 Answers 1. Suppose that A, B, and C are events, and you know that P(A) = 0.72

P(A ∪ B) = 0.78

P(B) = 0.50

P(B ∪ C) = 0.84

P(C) = 0.64

P(A ∪ C) = 0.91

P(A ∪ B ∪ C) = 0.95

What is P(ABC)? Answer : We compute P(AB) = P(A) + P(B) − P(A ∪ B) = 0.44 P(AC) = P(A) + P(C) − P(A ∪ C) = 0.45 P(BC) = P(B) + P(C) − P(B ∪ C) = 0.30

Then P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(AB) − P(AC) − P(BC) + P(ABC) 0.95 = 0.67 + P(ABC) P(ABC) = 0.28

2. Suppose that E and F are events, and P(E) = 0.9 and P(F) = 0.8. Prove that P(EF) > 0.7. In general, prove that P(EF) > P(E) + P(F) − 1.

Answer :

In the particular case, we know that P(E) + P(F) − P(EF) = P(E ∪ F) 6 1, so 1.7 − P(EF) > 1, or 0.7 > P(EF). In general, we just have P(EF) = P(E) + P(F) − P(E ∪ F) > P(E) + P(F) − 1, because P(E ∪ F) 6 1. 3. Remember that in bridge, the entire deck of 52 cards is dealt to 4 players, so that each player has 13 cards. In class,   we saw that the probability that each player receives exactly 1 ace is

4 1,1,1,1

48 12,12,12,12  52 13,13,13,13

≈ 0.1055.

(a ) What is the probability that North has 2 aces, South has 0 aces, and East and West each have 1 ace? (b ) What is the probability that one of the 4 players has 2 aces, one has 0 aces, and the other two each have 1 ace? (c ) What is the probability that 2 of the 4 players each have 2 aces, and the other two players have no aces?

(d ) What is the probability that 1 of the players has 3 aces, one has 1 ace, and the other two players have no aces? (e ) What is the probability that 1 player has all 4 aces?

Answer : (a ) This is

48 11,13,12,12  52 13,13,13,13



48 11,11,13,13  52 13,13,13,13



4 2,0,1,1



=

1014 ≈ 0.0487. 20825

(b ) We take the probability from the previous problem, and multiply by the number of ways that one player can have 2 aces, one can have none, and the remaining two aces end with one in each hand. The multiplication principle tells us that this can happen in 12 1014 · 12 = 12168 ≈ 0.5843. ways, so the probability is 20825 20825 (c ) The probability that North and South each have 2 aces is 4 2,2,0,0



=

468 ≈ 0.0225. 20825

We need the number of ways to choose the 2 players with 2 aces, which is 468 2808 desired probability is 20825 · 6 = 20825 ≈ 0.1348. (d ) The probability that North has 3 aces and South has 1 is 4 3,1,0,0

48 10,12,13,13  52 13,13,13,13



 =

4 2



= 6, so the

286 ≈ 0.0137. 20825

The number of ways to choose the player with 1 ace and the player with 3 aces is 12, so 286 3432 the desired probability is 20825 · 12 = 20825 ≈ 0.1648. (e ) The probability that North has 4 aces is 4 4,0,0,0

48 9,13,13,13  52 13,13,13,13



 =

11 ≈ 0.0026. 4165

This needs to be multiplied by 4 to compute the desired answer, because any one of the 4 11 44 players could have all 4 aces. The desired answer is 4165 · 4 = 4165 ≈ 0.0106. Note that all 5 fractions (including the one that we computed in class) sum to 1, as they must. Note also that the most likely event by far is that one player has 2 aces, one has none, and the remaining two aces are distributed among the remaining two players. 4. The following intriguing problem is from the text: Compute the probability that a bridge hand is void in at least one suit. (In other words, compute the probability that all 13 cards in a hand  lie  in at most 3 suits.) The text goes on to point out that the correct

answer is not

4 3

39 13  52 13

.

Explain why this fraction does not give the correct answer, and then use the Inclusion{ Exclusion Principle to compute the correct answer.

Answer : Let E1 = {a hand is void in spades} E2 = {a hand is void in hearts} E3 = {a hand is void in diamonds} E4 = {a hand is void in clubs}

We need P(E1 ∪ E2 ∪ E3 ∪ E4 ). It is indeed true that P(E1 ) = P(E2 ) = P(E3 ) = P(E4 ) =

39 13  52 13



. But to compute P(E1 ∪

E2 ∪ E3 ∪ E4 ), we must apply the Inclusion{Exclusion Principle. We start by computing  39 P(E1 ) = P(E1 E2 ) = P(E1 E2 E3 ) =

13  52 13  26 13  52 13  13 13  52 13

P(E1 E2 E3 E4 ) = 0

Because each sum in the expansion consists of equal terms, we can simplify to get:         4 4 4 4 P(E1 ∪ E2 ∪ E3 ∪ E4 ) = P(E1 ) − P(E1 E2 ) + P(E1 E2 E3 ) − P(E1 E2 E3 E4 ) 1 2 3 4    6 26 4 13 4 39 1621364909 13 13 13 ≈ 0.051066. = 52 − 52 + 52 = 31750677980 13 13 13

The originally suggested incorrect answer gives error is extremely small.

17063919 333515525

≈ 0.051164. In other words, the

5. A standard cubical die is rolled, and then rolled a second time. What is the probability that the second roll is higher than the rst?

Answer :

One way to do this is to go through all 36 elements in the sample space, and count how many times the second roll is larger than the rst. The answer is 15, so the 5 probability is 15 = 12 ≈ 0.4167. 36 A slicker way to proceed is to note that P(second roll higher than rst) = P(second roll lower than rst) P(second roll higher than rst) + P(second roll lower than rst) = 1 − P(second roll equals rst)

But P(second roll equals rst) = 61 , so we have 2P(second roll higher than rst) = 5 Again, we see that the desired probability is 12 .

5 6

.

6. An instructor gives his probability class a list of 13 problems, and tells the class that the nal examination will consist of 7 of these problems, chosen at random. A student chooses to solve 9 of the problems, and ignore the other 4. (a ) What is the probability that all 7 of the questions on the nal exam appeared on the student's list of 9 problems? (b ) What is the probability that at least 5 of the 7 problems on the nal exam appear on the student's list?

Answer : (a ) Think of the sample space in terms of the professor's choice of the problems

to be on the exam. That approach tells us that the probability is 9 7  13 7



=

3 ≈ 0.0210. 143

An observation: you can instead think of the sample space in terms of the student's choices, in which case the sample space consists of the students 9 choices of problems. He must choose all 7 of the problems that are on the exam, so he is left with 2 choices of problems ( 6) 3 of the 6 remaining ones, yielding (132 ) = 143 as before. 9 (b ) This is 9 7



+

9 6

 4 1  13 7

+

9 5

 4 2

=

94 ≈ 0.6573 143

7. An urn contains n white balls and m black balls, where n and m are positive integers. (a ) If two balls are withdrawn at random without replacement, what is the probability that they are the same color? (b ) If one ball is withdrawn at random and then replaced in the urn, and then a second ball is withdrawn at random from the urn, what is the probability that the two balls have the same color? (c ) Show that for all values of n and m, the probability in part (b ) is always larger than the probability in part (a ). Note : When showing that an inequality is true, you must begin with a known result and end with the desired inequality. It is incorrect to begin with the conclusion and work backwards until arriving at a true inequality.

Answer : (a ) This is n 2



+

m+n 2

m 2

 =

n 2



+

m+n 2

m 2



, but we need to simplify the answer for part (c ):

n(n−1) + m(m−1) 2 2 (m+n)(m+n−1) 2

=

n(n − 1) + m(m − 1) n2 − n + m2 − m = . (m + n)(m + n − 1) (m + n)2 − (m + n)

(b ) The probability that both balls are white is

n2 and the probability that both (n + m)2

m2 , so the probability that both are the same color is the sum balls are black is (n + m)2 n2 + m2 (n + m)2

(c ) We are asked to prove that for all positive values of m and n, n2 − n + m2 − m n2 + m2 > . (n + m)2 (m + n)2 − (m + n)

To do this, we'll show separately that: n2 n2 − n > (n + m)2 (m + n)2 − (m + n) m2 − m m2 > . (n + m)2 (m + n)2 − (m + n)

(1) (2)

I worked forward until I saw something that I knew was true, and then derived the desired result: 0 > −m

Add n2 + nm − n to both sides: n2 + nm − n > n2 + nm − n − m

Factor: n(n + m − 1) > (n + m)(n − 1)

Divide by (n + m − 1)(n + m): n n−1 > n+m n+m−1

Multiply by

n : n+m n2 n(n − 1) > 2 (n + m) (n + m)(n + m − 1) n2 − n = (n + m)2 − (n + m)

This proves (1), and then (2) follows by switching n and m. We then get the desired result by adding those two inequalities.