Chapter 3. Probability

70 Chapter 3: ProbabilityIn Exercises 9-13, use the following table, which summarizes blood groups and Rh types for100 typical people. These values may vary in different regions according to the ethnicity of thepopulation.Rh TypeO A B AB TotalPositive 39 35 8 4 86Negative 6 5 2 1 14Total 45 40 10 5 1009. Blood Groups and Types P(both of two randomly selected have O blood)a. with first selection replaced back into same data poolP(both O)= P(first O) ∗ P(second O from same pool)= 45/100 ∗ 45/100=0.450 ∗ 0.450= 0.203b. with first selection not replaced in the same data poolP(both O)= P(first O) ∗ P(second O from adjusted pool)= 45/100 ∗ 44/99=0.4500 ∗ 0.4444= 0.20010. Blood Groups and Types P(both of two randomly selected have Rh + blood)a. with first selection replaced back into same data poolP(both Rh + )= P(first Rh + ) ∗ P(second Rh + from same pool)= 86/100 ∗ 86/100=0.860 ∗ 0.860= 0.740b. with first selection not replaced in the same data poolP(both Rh + )= P(first Rh + ) ∗ P(second Rh + from adjusted pool)= 86/100 ∗ 85/99=0.8600 ∗ 0.8586= 0.73811. Blood Groups and Types P(three of three randomly selected have B blood)a. with first and second selections replaced back into same data poolP(all B)= P(first B) ∗ P(second B from same pool) ∗ P(third B from same pool)=10/100 ∗ 10/100 ∗ 10/100= 0.100 ∗ 0.100 ∗ 0.100= 0.00100b. with first and second selections not replaced in the same data poolP(all B)= P(first B) ∗ P(second B from adjusted pool) ∗ P(third B from adjusted pool)=10/100 ∗ 9/99 ∗ 8/98= 0.1000 ∗ 0.0909 ∗ 0.0816= 0.00074212. Blood Groups and Types P(four of four randomly selected have Rh – blood)a. with first, second, and third selections replaced back into same data poolP(all Rh – )= P(first Rh – ) ∗ P(second Rh – ) ∗ P(third Rh – ) ∗ P(fourth Rh – )=14/100 ∗ 14/100 ∗ 14/100 ∗ 14/100= 0.140 ∗ 0.140 ∗ 0.140 ∗ 0.140= 0.000384b. with first, second, and third selections not replaced in the same data poolP(all Rh – )= P(first Rh – ) ∗ P(second Rh – ) ∗ P(third Rh – ) ∗ P(fourth Rh – )=14/100 ∗ 13/99 ∗ 12/98 ∗ 11/97= 0.1400 ∗ 0.1313 ∗ 0.1224 ∗ 0.1134= 0.00025513. Blood Groups and TypesP(10 randomly selected will have all have Type A blood)= 0.400 10 = 0.00010514. Wearing Hunting Orange6 out of 123 injured hunters were wearing hunting orange, 2 to be picked at randoma. P(two selected at random with replacement)= P(first) ∗ P(second)= 6/123 ∗ 6/123=0.0488 2 = 0.00238b. P(two selected at random without replacement)= P(first) ∗ P(second)= 6/123 ∗ 5/122=0.0488 ∗ 0.0410= 0.00200c. In this case, the population size is relatively small so the sampling without replacement would betterrepresent the actual probability of randomly selecting two injured hunters who were wearing hunter orangefrom a population of 123 injured hunters.