Answer:
The probability that the family has at least one girl, given that they have at least one boy 0.80.
Step-by-step explanation:
Construct the sample space for all the 5 children as follows:
S = {GGGGG, GGGGB, GGGBB, GGBBB, GBBBB and BBBBB}
There are a total of 6 outcomes.
First compute the probability of at least 1 boy.
Favorable outcomes : {GGGGB, GGGBB, GGBBB, GBBBB and BBBBB} = 5
Then the probability of at least 1 boy is,
![P(At\ least\ 1\ boy)=\frac{5}{6}](https://tex.z-dn.net/?f=P%28At%5C%20least%5C%201%5C%20boy%29%3D%5Cfrac%7B5%7D%7B6%7D)
Of these 5 samples the sample space for at least 1 girl is:
{GGGGB, GGGBB, GGBBB and GBBBB} = 4
Then the probability of at least 1 girl given at least 1 boy is,
![P(At\ least\ 1\ girl|At\ least\ 1\ boy)=\frac{4}{5}=0.80](https://tex.z-dn.net/?f=P%28At%5C%20least%5C%201%5C%20girl%7CAt%5C%20least%5C%201%5C%20boy%29%3D%5Cfrac%7B4%7D%7B5%7D%3D0.80)
Thus, the probability of at least 1 girl given at least 1 boy is, 0.80.