hope4dabest wrote:
i guess i got it..chances are 50% that is 1/2 and as he has 5 children so probability is 1/32 but still plz do explain it...
Sorry for late reply. The probability of having one child affected for that couple will be 1/2. So, the probability of having all 5 children affected will be 1/2 * 1/2 * 1/2 * 1/2 * 1/2 which is 1/32.
That is the application of binomial equation.
Let's say this couple have 3 children and what is the probability of having 2 children unaffected and one children affected?
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| hope4dabest
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unaffected---1/2*1/2=1/4
affected--1/2
did i get it?????
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| drgho
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hope4dabest wrote:
unaffected---1/2*1/2=1/4
affected--1/2
did i get it?????
 ...only one probability... not two probabilities separated.
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| hope4dabest
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drgho wrote: ...only one probability... not two probabilities separated.
but there r 3 children in this case and chance of expressing this gene is 1/2 and that also means not expressing is the left out that is 1/2....and as no of children are independent events so out of 3 children, probability 0f 2 children for not affected wont be multiplying 2 times and and for affected.........
i know i am in deep shit...plzzz explain again???
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| hope4dabest
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or do u mean probability for 3 children of being affected or unaffected wil be the same that is 1/2*1/2*1/2
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| drgho
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hope4dabest wrote:
or do u mean probability for 3 children of being affected or unaffected wil be the same that is 1/2*1/2*1/2
yep..it is 1/8 but...there are also three combinations to have one affected and two unaffected. Out of three children named a, b and c. It is possible that Abc or aBc or abC (capital letter for affected one). So the probability is 3/8.
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| hope4dabest
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ok..thank u sooo much....i will try to look some more questions of these types....
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