STAT 1000 Lecture Notes - Lecture 25: Normal Distribution, Sample Space, Random Variable

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P(o) + p(a) + p(b) + p(ab) = 0. 48 + 0. 23 + 0. 19 + 0. 10 = 1. The probability of not having blood type a is: P(ac) = 1 p(a) = 1 0. 23 = 0. 77. We could also find it as follows: P(ac) = p(o or b or ab) = p(o) + p(b) + p(ab) = 0. 48 + 0. 19 + 0. 10 = 0. 77. The blood types of any two people are independent. the probability that the first person has blood type a and the second person has blood type o is: P(first has type a and second has type o) = p(first has type a)p(second has type. If the order doesn"t matter and we just want to know the probability that one of the two people has blood type a and the other has blood type o, then: P(one has type a and one has type o) = p(first has type a)p(second has type o) +

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