chapter 1

Chapter 2: Probability54. P(A 1 ) = .22, P(A 2 ) = .25, P(A 3 ) = .28, P(A 1 ∩ A 2 ) = .11, P(A 1 ∩ A 3 ) = .05, P(A 2 ∩ A 3 ) = .07,P(A 1 ∩ A 2 ∩ A 3 ) = .01P(A1∩ A2) .11a. P(A 2 ⏐A 1 ) = = = . 50P(A ) .221P(A1∩ A2∩ A3).01b. P(A 2 ∩ A 3 ⏐A 1 ) = = = . 0455P(A ) .221c.P(A2∪ A | A ) =3P(A1∩ A=21) + P(AP[A1∩ ( A2∪ A3)]=P(A )111∩ A3)− P(A1∩ AP(A )P[(A21∩ A2) ∪ ( AP(A )11∩ A∩ A3).15= = .682.223)]P(A1∩ A2∩ A3).01d. P ( A1 ∩ A2∩ A3| A1∪ A2∪ A3) == = . 0189P(A ∪ A ∪ A ) .53This is the probability of being awarded all three projects given that at least one projectwas awarded.12355.2 × 12 × 1a. P(A B) = P(B|A)•P(A) = × = . 01114 × 36 × 5b. P(two other H’s next to their wives | J and M together in the middle)P[(H −W.orW . − H ) and(J − M.or.M − J)and(H −W.orW .P(J − M.or.M − J.in.the.middle)numerator =denominator =4 × 1×2 × 1×2 × 1 16=6 × 5×4 × 3×2 × 1 6!4 × 3×2 × 1×2 × 1 48=6 × 5 × 4 × 3×2 × 1 6!16 1 = .48 3so the desired probability =− H)]72