doc179
Forum Guru
Topics: 67
Posts: 1,217
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the ans for the first Q seems like c. The possibility that a child will be affected is exactly the same as it being unaffected. I think I did it the hard way, by writing out all possibilities. If A is affected and X is not affected, the possibilities are:
AAA
XAA
AXA
AAX
XXA
XAX
AXX
XXX
Eight possibilities. According to this, the chance of
All children being affected is 1/8
One child being affected is 3/8
Two children being affected is 3/8
All children being affected 1/8.
The answer is C if the question means what is the probability of EXACTLY one child being affected. The probability of ATLEAST one child being affected is 7/8 (E).
Now I need someone to help me interpret this question right. Exactly one child or atleast one child?
Will post for the second part later, if I can..........
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| doc179
Forum Guru
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Posts: 1,217
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1/2+1/2+1/2 = 3/8???
32/16 = 1/2 ???
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| sarika
Forum Guru
Topics: 195
Posts: 1,200
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Thats exactly what i was saying....thats how wrong i am 
My maths is As terrible as my medicine.
Edited by sarika on 02/21/06 - 12:05 AM
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| robin082006
Forum Hero
Topics: 471
Posts: 5,125
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because AD, every child has 1/2 chance to have disease and 1/2 chance normal.
so we think about coin tossing
for the first question, doc179 is right, ans is 3/8
for the second question, ans is E. 15/16
Gardner is AD disease, each child has 1/2 chance to have disease, the probability that all 4 children are normal is 1/2*1/2*1/2*1/2= 1/16, so the probability that at least one child has disease is 1-1/6= 15/16.
___________________
The Key to Succeed is Patience.
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| robin082006
Forum Hero
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more explanation about the second question
all 4 children are normal = 1/16
exactly 1 child with disease = 4/16
exactly 2 children with disease = 6/16
exactly 3 children with disease = 4/16
all 4 children with disases = 1/16
___________________
The Key to Succeed is Patience.
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