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18 ≈ 1. 54. 2 Chapter 4 55. 2 e + e 2 2 56. The number of people in a random collection of size n that have the same birthday as yourself is approximately Poisson distributed with mean n/365. Hence, the probability that at least one person has the same birthday as you is approximately 1 − e−n/365. Now, e−x = 1/2 when x = log(2). Thus, 1 − e−n/365 ≥ 1/2 when n/365 ≥ log(2). That is, there must be at least 365 log(2) people. 57. (a) 1 − e−3 − 3e−3 − e−3 32 17 = 1 − e−3 2 2 17 1 − e−3 P{ X ≥ 3} 2 = (b) P{X ≥ 3X ≥ 1} = P{ X ≥ 1} 1 − e−3 59.

4) 25 53. Let W and F be the events that component 1 works and that the system functions. P(WF) = 55. P{Boy, F} = P(WF ) P(W ) 1/ 2 = = c P( F ) 1 − P( F ) 1 − (1 / 2) n −1 4 16 + x P{Boy) = so independence ⇒ 4 = 10 16 + x P{F} = 10 16 + x 10 ⋅ 10 ⇒ 4x = 36 or x= 9. 16 + x A direct check now shows that 9 sophomore girls (which the above shows is necessary) is also sufficient for independence of sex and class. 56. P{new} = ∑ P{new i 30 type i}pi = ∑ (1 − p ) i n −1 pi i Chapter 3 57. (a) 2p(1 − p) 3 (b)   p 2 (1 − p)  2 (c) P{up on firstup 1 after 3} = P{up first, up 1 after 3}/[3p2(1 − p)] = p2p(1 − p)/[3p2(1 − p)] = 2/3.

Chapter 3 39 Theoretical Exercises 1. P(ABA) = 2. If A ⊂ B P( AB ) P( AB ) ≥ = P(ABA ∪ B) P( A) P( A ∪ B) P( A) , P(ABc) = 0, P( B) P(AB) = 3. P(BA) = 1, P(BAc) = P( BAc ) P( Ac ) Let F be the event that a first born is chosen. Also, let Si be the event that the family chosen in method a is of size i. Pa(F) = 1 ni ∑ P( F S ) P( S ) = ∑ i m i i i Pb(F) = i m i ini ∑ Thus, we must show that ∑ in ∑ n / i ≥ m i i i 2 i or, equivalently, ∑ in ∑ n i i j /j≥ j ∑n ∑n i i j j or, i ∑∑ j n n ≥ ∑∑ n n i i≠ j j i j i≠ j Considering the coefficients of the term ninj, shows that it is sufficient to establish that i j + ≥2 j i or equivalently i2 + j2 ≥ 2ij which follows since (i − j)2 ≥ 0.

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