chapter 1

Chapter 2: Probability47.P(A ∩ B)P(A).25.50a. P(B⏐A) = = = . 50P(A ∩ B′)P(A).25.50b. P(B′⏐A) = = = . 50P(A ∩ B)P(B).25.40c. P(A⏐B) = = = . 6125P(A′∩ B)P(B).15.40d. P(A′⏐B) = = = . 3875P[A ∩ ( A ∪ B)]P(A ∪ B).50.65e. P(A⏐A∪B) = = = . 769248.P(A1∩ A2) .06a. P(A 2 ⏐A 1 ) = = = . 50P(A ) .12.01b. P(A 1 ∩ A 2 ∩ A 3 ⏐A 1 ) = = . 0833.121c. We want P[(exactly one) ⏐ (at least one)].P(at least one) = P(A 1 ∪ A 2 ∪ A 3 )= .12 + .07 + .05 - .06 - .03 - .02 + .01 = .14Also notice that the intersection of the two events is just the 1 st event, since “exactly one”is totally contained in “at least one.”.04 + .01So P[(exactly one) ⏐ (at least one)]= = . 357169.14d. The pieces of this equation can be found in your answers to exercise 26 (section 2.2):P(A1∩ A2∩ A′3) .05P ( A′3| A1∩ A2)== = .833P(A ∩ A ) .0612