SOLUTION: a family contain 8 childrens half of which are boys and half of which are girls if 5 childrens are selected from the family than find the probability distribution of no. of boys

There are 4 boys and 4 girls. Therefore:

in a random selection of 5 children, there are either x = exactly 1 boy, x = 
exactly 2 boys, x = exactly 3 boys, or x = exactly 4 boys.

Number of ways to have:

x = exactly 1 boy:  Choose the 1 boy 4C1=4 ways and all 4 girls 4C4=1 way. That's 4∙1 = 4 ways.
x = exactly 2 boys:  Choose the 2 boys 4C2=6 ways and the 3 girls 4C3=4 ways. That's 6∙4 = 24 ways.
x = exactly 3 boys:  Choose the 3 boys 4C3=4 ways and the 2 girls 4C2=6 ways. That's 4∙6 = 24 ways.
x = exactly 4 boys:  Choose all 4 boys 4C4=1 way and the 1 girl 4C1=4 ways. That's 1∙4 = 4 ways. 

So there are 4+24+24+4 = 56 ways to choose 5 from the 8.

[As a check on the 56, let's calculate the number of ways we can choose 5

children from the 8. --> 8 children CHOOSE 5 = 8C5 = 56.  So it checks!]

To find the probability of each, we divide the number of ways by

      x    f(x)  x∙f(x)  
      1     4      4
      2    24     48
      3    24     72 
      4     4     16     
     ------------------
    Σ      56    140

We calculate the mean

μ = Σ[x∙f(x)]/Σ[f(x)] = 140/56 = 2.5   

      x    f(x)  x-μ   (x-μ)²   f(x)∙(x-μ)² 
      1     4   -1.5    2.25        9  
      2    24   -0.5    0.25        6
      3    24    0.5    0.25        6 
      4     4    1.5    2.25        9
     --------------------------------------
    Σ      56                      30

This is the population, not a sample.  So we divide 30/56 and reduce to
15/28.  

Variance = σ² = 15/28 = 0.5357142957
S.D. = σ = √(σ²) = √(0.5357142957) = 0.7319250547

Edwin