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abhupathi

Guest

probability - "4 children"

hi,

dont understand this at all.

Abena and Kofi want to have 4 children, but the probability that a child is going to have a disease is 1/4

work out the probability that at most one of their 4 children will have a disease.

i have no clue what to do here, i really don't...any help?

thanks. answer is 0.738, but i thought it should be just 1/4 anyway...help?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: probability - "4 children"

hi abhupathi

It could happen that none or one or  two or three or even all four have the disease.

This is an example of what is called the 'Binomial Distribution'.  It is worked out like this:

P(none have the disease) = 3/4  x  3/4  x 3/4  x 3/4  (As each must 'not get it')  = 81/256

P(a particular child has it) = 1/4 x 3/4  x 3/4  x 3/4    ( one has but three have not!)

but 'a particular child' means one of four so

P(any one child has it) = 4 x (1/4  x 3/4  x 3/4  x 3/4)  = 27/64

So now it is possible to answer the question.

P(at most one) = P(none) + P(any one has it) =  81/256  +  27/64 =   0.73828125...

I have added a diagram which may help.  Start at the left and follow a route going right.  I have shown one way in which a single child may get the disease.  You should be able to find three other routes for this outcome.  And I have shown the way in which no child gets the disease.

You can see that there are 16 posssible outcomes.  You could try to work out the probabilities for each route.  When you add up all 16 answers you should get exactly 1.0000000

Post back if you want to know more about the Binomial Distribution'.

Bob

Last edited by Bob (2010-11-29 08:32:08)

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