# Homework Help: Game show there are 10 contestants of which 6 are female.

1. Nov 24, 2011

### AllenHe

1. The problem statement, all variables and given/known data
At the start of a gameshow there are 10 contestants of which 6 are female. In each round of
the game, one contestant is eliminated. All of the contestants have the same chance of
progressing to the next round each time.

Given that the first contestant to be eliminated is male, find the probability that the next
two contestants to be eliminated are both female.

2. Relevant equations
p(next to be female|male)=(next to be female n male)/(male)

3. The attempt at a solution
I did (4/10)*(6/9)*(5/8)
But when I checked the answer, it was (6/9)*(5/8).
Why?

2. Nov 24, 2011

### Dick

Re: probability

This is probably just a language issue. The problem says "Given that the first contestant to be eliminated is male". "Given" means that you assume the male is eliminated before you start calculating probabilities. It's called a conditional probability.

3. Nov 24, 2011

### AllenHe

Re: probability

Ya, that's what I did. But how come I didn't get the same answer?

4. Nov 24, 2011

### AllenHe

Re: probability

Or is there something wrong with my equation?

5. Nov 24, 2011

### Dick

Re: probability

"Given" means something has already taken place and you don't factor in that probability. There's nothing wrong with the equation, but "given" means skip the 4/10 factor.

6. Nov 24, 2011

### HallsofIvy

Re: probability

Originally, there were 10 contestants and 6 were female. Given that the first to be eliminated was male, there are now 9 contestants and 6 are female. What is the probability that the one eliminated now will be female? If that happens then there will be 8 contestants, 5 of them female. What is the probability that the one eliminated now will be female?

7. Nov 24, 2011

### AllenHe

Re: probability

But how can I use the equation :

P(A|B)=P(AnB)/P(B)

to get the answer?Or is it not possible, because the total number of people decreases?

8. Nov 24, 2011

### Ray Vickson

Re: probability

Sometimes we use that equation in reverse. When we know P(B) and P(A|B) we can use the equation to get P(AnB). That is what is happening in this problem.

RGV

9. Nov 24, 2011

### AllenHe

Re: probability

so what's the value of P(AnB)?

10. Nov 25, 2011

### Ray Vickson

Re: probability

Tell me what are A and B. You brought up the AnB, and I just responded to your question.

RGV

11. Nov 25, 2011

### AllenHe

Re: probability

(4/10)*(6/9)*(5/8)
But it's not correct.

12. Nov 25, 2011

### Ray Vickson

Re: probability

Of course not. As has already been explained clearly to you, there should be no 4/10 factor.

RGV

13. Nov 25, 2011

### ArcanaNoir

Re: probability

the probability of two independent events A and B: P(AnB) is the probability of A times the probability of B. Thats two things, not three things. So, not (4/10)*(6/9)*(5/8). Just (6/9)*(5/8)

Show that $\lim_{n\to\infty}(1+\frac{r}{n})^n=e^r$