Solutions HWB05

3.8a: Pr(Both Watch) = Pr(Husband Watches)Pr(WifeWatches | Husband Watches) = (.20)(.25) = .05

3.8b: Pr(Husband Watches or Wife Watches) = Pr(Husband Watches) + Pr(Wife Watches) – Pr(Both Watch) = .20 + .08 - .05 = .23

3.8c: Pr(Neither Watches) = 1 – Pr(At least one Watches) = 1 - .23 = .77

3.8d: Pr(Husand Watches | Wife Watches) = Pr(Both Watch)/Pr(Wife Watches) = .05/.08 = .625

3.10: Pr(E1 or E2) = Pr(E1) + Pr(E2) – Pr(Both bad) = 3/100 + 3/100 – (3/100)(2/99) = 588/9900

or

Pr(E1 or E2) = 1 – Pr(Both Good) = 1 – (97/100)(96/99) = 1 – 9312/9900 = 588/9900.

3.12: Pr(at least one ace in 4 rolls) = 1 – Pr(No Aces in 4 rolls) = 1 – (5/6)^4 = .5177

Pr(at least one SnakeEyes in 24 rolls) = 1 – Pr(No SnakeEyes in 24 rolls) = 1 – (35/36)^24 = .4914

3.70: Various solutions.

Experiment: Pick a card from a 52-card deck.

Events of “Get Heart” and “Get Spade” are disjoint: Pr(Card is both Heart and Spade) = 0.

The event of “Get Heart” and “Get Ace” are independent:

Pr(Heart | Ace) =1/4.

Pr(Heart) = 13/52 = 1/4.