1. ## Flipping 10 coins

Say that a coin is flipped 10 times. What is the probability that the first 3 flips are all heads provided that there are an even number of heads and tails all together?

- I understand that if we simply flip 3 coins, the probability that 3 heads appear is 1/6. How do we take into account the fact that 5 of 10 are definitely heads??

-Thanks

2. Originally Posted by ccdelia7
- I understand that if we simply flip 3 coins, the probability that 3 heads appear is 1/6.
-Thanks
1/8

RonL

3. Originally Posted by ccdelia7
provided that there are an even number of heads and tails all together?
This conveys no information all the 10 flips are either heads of tails, and 10 is an even number.

Do you mean:

provided that there are an even number of heads

RonL

4. Originally Posted by ccdelia7
Say that a coin is flipped 10 times. What is the probability that the first 3 flips are all heads provided that there are an even number of heads and tails all together?
- I understand that if we simply flip 3 coins, the probability that 3 heads appear is 1/6.
First lets correct the mistake (typo?) above.
Flipping a coin three times gives 8 possible outcomes only one of which is “HHH”.
Therefore the probability of “HHH” is 1/8 not 1/6.

Thus the same idea gives an approach to the question.
If in ten flips we have an even number of heads, 0 2 4 6 8 10, then we have an even number of tails.
In ten flips that can happen in $\displaystyle \sum\limits_{k = 0}^5 \binom{10}{2k}$ ways.

Now we must find out how many of those begin the string with “HHH…”.
If we already have three H’s then in the next seven places we must have 1, 3, 5, or 7 H’s to have an even number of heads and tails all together.
The number of ways that can happens is $\displaystyle \sum\limits_{k = 1}^4 \binom{7}{2k-1}$.

Simply divide the two.

5. I apologize! I meant to write equal. There is an equal number of heads and tails. (i.e. - there are 5 heads and 5 tails ) I am sorry for misleading everyone! Thought the problem you have solved is a quite interesting one!!

6. That makes it easy: $\displaystyle \frac {\binom {7}{2}} {\binom {10}{5}}$

7. yea - I am sorry! It's much easier than it looked, I guess!

8. now - I am not seeing why exactly we can rewrite this event as a binomial cofficient?? Can you explain? i can kind of see where you are going with it, but it's still a bit vague. Thanks!