Solutions to Exercises


    1. {Gl, G2, G3, G4, G5, Yl, Y2, Y3}
    2. \(\dfrac{5}{8}\)
    3. \(\dfrac{2}{3}\)
    4. \(\dfrac{2}{8}\)
    5. \(\dfrac{6}{8}\)
    6. No


    1. {(HHH) , (HHT) , (HTH) , (HTT) , (THH) , (THT) , (TTH) , (TTT)}
    2. \(\dfrac{4}{8}\)
    3. Yes

  3. 0


    1. 0
    2. 0
    3. 0.63


    1. \(\dfrac{43}{215}\)
    2. \(\dfrac{120}{215}\)
    3. \(\dfrac{20}{215}\)
    4. \(\dfrac{12}{172}\)
    5. \(\dfrac{115}{215}\)


    1. iii
    2. i
    3. iv
    4. ii


    1.  P(H or G) = P(H) + P(G) - P(H and G) = 0.26 + 0.43 - 0.14 = 0.55
    2. P( NOT (H and G) ) = 1 - P(H and G) = 1 - 0.14 = 0.86
    3. P( NOT (H or G) ) = 1 - P(H or G) = 1 - 0.55 = 0.45


    1. P (Type O or Rh-) = P(Type O) + P(Rh-) - P(Type O and Rh-)

      0.52 = 0.43 + 0.15 - P(Type O and Rh-); solve to find P(Type O and Rh-) = 0.06

      6% of people have type O Rh— blood

    2. P( NOT (Type O and Rh-) ) = 1 - P(Type O and Rh-) = 1 - 0.06 = 0.94

      94% of people do not have type O Rh— blood


    1. P(R or F) = P(R) + P(F) - P(R and F) = 0.72 + 0.46 - 0.32 = 0.86
    2. P( Neither R nor F ) = 1 - P(R or F) = 1 - 0.86 = 0.14


    1. P(D and E) = P(D I E)P(E) = (0.20)(0.40) = 0.08
    2. P(E I D) = P(D and E) / P(D) = 0.08/0.10 = 0.80
    3. P(D or E) = P(D) + P(E) - P(D and E) = 0.10 + 0.40 - 0.08 = 0.42
    4. Not Independent: P(D I E) = 0.20 which does not equal P(D) = .10
    5. Not Mutually Exclusive: P(D and E) = 0.08 ; if they were mutually exclusive then we would need to have P(D and E) = 0, which is not true here.