By Sheldon Ross

**Read or Download A First Course In Probability (Solution Manual) PDF**

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**Additional resources for A First Course In Probability (Solution Manual)**

**Sample text**

Chapter 3 39 Theoretical Exercises 1. P(ABA) = 2. If A ⊂ B P( AB ) P( AB ) ≥ = P(ABA ∪ B) P( A) P( A ∪ B) P( A) , P(ABc) = 0, P( B) P(AB) = 3. P(BA) = 1, P(BAc) = 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.

Assume n > 1. 2 (a) n (b) Conditioning on whether the man of couple j sits next to the woman of couple i gives the 1 1 n−2 2 2n − 3 result: + = n − 1 n − 1 n − 1 n − 1 (n − 1) 2 (c) e−2 68. exp(−10e−5} 69. 2212 54 Chapter 4 70. e−λt + (1 − e−λt)p 71. 26 (a) 38 5 3 26 12 (b) 38 38 72. 4) 3 73. Let N be the number of games played. 8125 74. 2 (a) 3 5 5 3 6 2 7 8 8 2 1 8 2 1 8 2 1 2 (b) + + + 5 3 3 6 3 3 7 3 3 3 5 5 2 1 (c) 4 3 3 6 2 (d) 4 3 5 1 3 2 76.

E[Y] = E[X/σ − µ/σ] = 1 σ E[ X ] − µ/σ = µ/σ − µ/σ = 0 Var(Y) = (1/σ)2 Var(X) = σ2/σ2 = 1. n 10. E[1/(X + 1)] = i =0 n = ∑ i =0 = = 1 n! i! p (1 − p) i n −i n! (i + 1)! n n + 1 1 (n + 1) p ∑ i + 1 p 1 ( n + 1) p ∑ i +1 (1 − p) n − i i =0 n +1 n + 1 j n +1− j p (1 − p ) j j =1 n + 1 0 1 n +1− 0 1 − 0 p (1 − p) (n + 1) p 1 = [1 − (1 − p ) n +1 ] (n + 1) p = 11. For any given arrangement of k successes and n − k failures: P{arrangementtotal of k successes} = 12.