robin082006
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C
Assume her 3 children named X, Y, Z
Each child has 1/2 chance of disease and 1/2 chance free of disease.
The probability that one of the three children will be affected= X with disease but other 2 without disease + Y with disease but other 2 without disease + Z with disease but other 2 without disease = 1/2*1/2*1/2 + 1/2*1/2*1/2 +1/2*1/2*1/2=3/8
All 3 have disease = 1/2*1/2*1/2=1/8
All 3 free of disease = 1/2*1/2*1/2=1/8
2 of 3 have disease 1-3/8-1/8-1/8=3/8
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| mjl1717
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i multiplied 1/2 to the 3rd power to get 1/8 comments please.
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| Diego Casali
Forum Elite
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I agree with robin082006.
In effect, 1/8 is only the probability that the first child will be affected.
We have to consider also the other two possibilities!
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| nadiabarati
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I don't understand it!!!
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| nadiabarati
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Hey robin. I've been always appreciating your explanations. Thanks . My exam is in two days. Good luck to you. Wish you the best.
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| robin082006
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In Two days?
Good luck to you.
Give yor experience after the exam.
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| Diego Casali
Forum Elite
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Good luck, nadiabarati.
Best wishes.
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