Explain the calculation for this simple probability problem (ANSWER GIVEN)?

Question

What is the probability of having 3 girls and 3 boys?

If you do out every combination of 6 children:
BBBBBB
BBBBBG
BBBBGB
BBBBGG
BBBGBB etc..., you will find that there are 64 combinations, and 20 of them have 3 girls and 3 boys. (answer=20/64)
What is the mathematical equation to simplify this, as writing that out would take 15 minutes?

Please show the worked problem for the probability of 3 girls and 3 boys, and also for the probability of at least 4 girls. (answer for that one is 22/64)

Thank you

Answer 1

(6C3)(3C3)/2^6
20/64


2) at LEAST 4 girls

[(6C4)(2C2) + (6C5)(1C1) + (6C6)(0C0)]/2^6

22/64

i hope this helped,,,try examining the answer and im sure you'll know how this worked out...

the 2^6 at the end is because the 2 refers to boy or girl,,,,and the ^6 refers to 6 separate events!

Answer 2

Prob(3B and 3G) = (6C3)(1/2)³(1/2)³ = 20/(8*8) = 20/64
_____________

Because 3 girls and 3 boys is in the middle the probability of more than 3 girls is the same as the probability of less than 3 girls. Each will be one-half of the complement of the probability of exactly three girls.

P(G≥4) = [1 - P(3B 3B)]/2 = (1 - 20/64)/2 = (44/64)/2 = 22/64

If you don't see that we can work it out directly.

P(G≥4) = P(4G 2B) + P(5G 1B) + P(6G)

= (6C4)(1/2)^4*(1/2)² + (6C5)(1/2)^5*(1/2) + (6C6)(1/2)^6

= [(6C4) + (6C5) + (6C6)](1/2)^6 = (15 + 6 + 1)/64 = 22/64