STAT 1000 Lecture Notes - Lecture 25: Binomial Distribution, Environment And Climate Change Canada, Random Variable

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Notice that the probability of each outcome is between 0 and 1. 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: = (0. 23)(0. 48) = 0. 1104. If the order does(cid:374)"t (cid:373)atter a(cid:374)d (cid:449)e just (cid:449)a(cid:374)t to k(cid:374)o(cid:449) the pro(cid:271)a(cid:271)ility that o(cid:374)e of the t(cid:449)o people has blood type a and the other has blood type o, then:

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