When coin 1 is flipped, it lands on heads with probability .4; when coin 2 is flipped, it lands on heads with probability .7. One of these coins is randomly chosen and flipped 10 times.
(a) What is the probability that the coin lands on heads on exactly 7 of the 10 flips?
(b) Given that the first of these 10 flips lands heads, what is the conditional probability that exactly 7 of the 10 flips land on heads?
I know how to to do by finding the probability of coin 1 being chosen then finding the probability it gets exactly 7 heads by using a bernoulli sum and adding that probability to the probability of coin 2 being chosen and doing another bernoulli sum for that coin, but is there a better or more efficient way of calculating the probability of part a instead of doing it the long way like I did?
and part b does not make sense to me because if it's given that the first 10 flips are heads how can you find the probability that exactly 7 of the 10 landed on heads if it's given that they all landed on heads?
Best Answer
As stressed in the comments, the extra information available in part b) is that fact that the first toss gives you Heads. Conceptually, the problem is "how do you use the new information?". Informally, the fact that you see $H$ on the first trial increases the probability that you have coin #$2$. But by how much?
To answer that, suppose you toss each coin ten times. You'll see $4$ Heads from coin #$1$ and $7$ from coin #$2$. Thus you'll see $11$ Head all in all, $7$ of which come from coin #$2$. Thus, if all you are told was that you saw Heads, the probability that it came from coin #2 is $\frac 7{11}$.
Accordingly, you can now redo part a), looking for exactly $6$ Heads out of the next $9$ tosses where you have coin #$1$ with probability $\frac 4{11}$ and coin #$2$ with probability $\frac 7{11}$.