Find the Probability

Question

A study determined that 5% of children under 18 years of age live with their father only. Find the probability that at most 2 individuals selected at random from 14 children under 18 years of age lived with their father only.

Answer 1

I think the question isn't well stated.  I think it should say:

 

"Find the probability that at most 2 individuals out of 14 children selected at random under 18 years of age lived with their father only."

 

Otherwise it doesn't really make sense.

 

OK...  When they say "at most 2" it means that it could be 0, 1 or 2.  Each of these is "mutually exclusive", which means that it can't be more than one number at the same time.  That means you can find the probabilities of 0 children living with their father only, 1 and 2, and add the three probabilities together.

 

So when selecting n items, the probability that r of them will have a specific attribute is

 

                                 n!

( P_yes^r )( P_no^(n-r) ) --------

                              r!(n-r)

 

P_yes is the probability that one item has the attribute.

P_no is the probability that one item does NOT have the attribute.

 

In this case, P_yes = 0.05 and P_no = 0.95.

 

So the probability that NONE of the children live with their father only is:

 

(0.05)⁰(0.95)¹⁴ * 14! / (0! * 14!) = 0.488

 

The probability that ONE of the children live with their father only is:

 

(0.05)¹(0.95)¹³ * 14! / (1! * 13!) = 0.359

 

The probability that TWO of the children live with their father only is:

 

(0.05)²(0.95)¹² * 14! / (2! * 12!) = 0.123

 

Adding the three probabilities together results in 0.970 or 97%