Answer:
- 0: 0.7056
- 1: 0.2688
- 2: 0.0256
Step-by-step explanation:
When events are independent, the probability of some sequence of them is the product of the probabilities of the individual events in that sequence.
The probability of a child having spina bifida is 16% = 0.16, so the probability that the child will not have the condition is 1 - 0.16 = 0.84. The probability that 0 of 2 children will have spina bifida is ...
p(0 for 2) = p(0 for 1)×p(0 for 1) = 0.84×0.84 = 0.7056
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There are two ways that 1 of 2 children can have spina bifida: either the first one does, or the second one does. These are mutually exclusive conditions, so their probabilities add:
p(1 for 2) = p(1 for 1)×p(0 for 1) +p(0 for 1)×p(1 for 1) = 0.16×0.84 +0.84×0.16
p(1 for 2) = 0.2688
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There is one way both children can have spina bifida:
p(2 for 2) = p(1 for 1)×p(1 for 1) = 0.16×0.16 = 0.0256
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In summary, our probability distribution is ...
p(X=0) = 0.7056
p(X=1) = 0.2688
p(X=2) = 0.0256
