SOLUTION: A family has five children. Assume that the probability of having a boy is 0.5. write the term in the expansion of (b+g)^5 for each outcome described. then evaluate each probabilit

Algebra ->  Permutations -> SOLUTION: A family has five children. Assume that the probability of having a boy is 0.5. write the term in the expansion of (b+g)^5 for each outcome described. then evaluate each probabilit      Log On


   

Question 251079: A family has five children. Assume that the probability of having a boy is 0.5. write the term in the expansion of (b+g)^5 for each outcome described. then evaluate each probability.
a. exactly 3 boys
b. exactly 4 boys
c. exactly 4 girls
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!


b=g=0.5, 

%28b%2Bg%29%5E5=b%5E5%2B5b%5E4g%2B10b%5E3g%5E2%2B10b%5E2g%5E3%2B5bg%5E4%2Bg%5E5

The 1st term b%5E5+=+%280.5%29%5E5+=+0.03125 is the probability of 5 boys

The 2nd term 5b%5E4g+=+5%280.5%29%5E4%280.5%29+=+0.15625 is the probability 
of 4 boys and 1 girl.

The 3rd term 10b%5E3g%5E2+=+10%280.5%29%5E3%280.5%29%5E2+=+0.3125 is the probability 
of 3 boys and 2 girls. 

The 4th term 10b%5E2g%5E3+=+10%280.5%29%5E2%280.5%29%5E3+=+0.3125 is the probability 
of 2 boys and 3 girls.

The 5th term 5b%5E4g+=+5%280.5%29%280.5%29%5E4+=+0.15625 is the probability 
of 1 boy and 4 girls.

The 6th term g%5E5+=+%280.5%29%5E5+=+0.03125 is the probability of 5 girls.

a. exactly 3 boys

That's the 3rd term 10b%5E3g%5E2+=+10%280.5%29%5E3%280.5%29%5E2+=+0.3125

b. exactly 4 boys

That's the 2nd term 5b%5E4g+=+5%280.5%29%5E4%280.5%29+=+0.15625

c. exactly 4 girls

That's the  5th term 5bg%5E4+=+5%280.5%29%280.5%29%5E4+=+0.15625

Edwin