# Can you verify my solutions to these probabilities, given a coin is flipped 10 times?

I'm fresh into probability and I think it's important to ask a lot of questions since it seems probability really challenges your intuition. I'm working on the following problem, and have found a solution for each, but I'd like to verify my understanding is correct.

I'm not entirely sure on part c (this IS the intersection, right? Is it simply just multiplying the two probabilities together, or is there something I'm missing? The formula below part c makes me nervous I'm missing something).

A coin is flipped 10 times.
a) What is the probability that there are exactly five heads and five tails?
b) What is the probability that the first three flips are heads?
c) What is the probability that there are exactly five heads and five tails AND the first three flips are heads?
d) What is the probability that there are exactly five heads and five tails OR the first three flips are heads?

Let $(S, P)$ be a sample space such that $$S = \{(a_1, a_2, a_3, ..., a_{10}) : a_1 \in \{H, T\}\}$$ and $P(S) = \frac{1}{2^{10}}$ for all $s \in S$.

a) By definition, $|S| = 1024$. Let $E$ be the event that we get 5 heads (so the other five flips are tails). Then $|E| = $$10 \choose 5, because we are choosing 5 of the 10 flips to be heads. Since all outcomes are equally likely, P(E) = \frac{|E|}{|S|} = \frac{252}{1024}. b) For each of the first three flps, there is a \frac{1}{2} chance that each one will be heads. By the multiplication principle, the probability that the first three flips are heads is \frac{1}{2^3} = \frac{1}{8}. c) From part a, we know that the probability of exactly five heads and five tails is \frac{252}{1024}. From part b, we know the probability that the first three flips are heads is \frac{1}{8}. So the probability of both is A \cap B = P(A) * P(B) = \frac{63}{2048} (defining A as the event of exactly five heads and tails, and B as the event of the first three flips being heads). d) We have the formula$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$. Using our A and B events from part c, P(A) = \frac{252}{1024}, P(B) = \frac{1}{8}, and P(A \cap B) = \frac{63}{2048}. So we have that$$P(A \cup B) = \frac{252}{1024} + \frac{1}{8} - \frac{63}{2048} = \frac{697}{2048}$$. • c) is not right, the events are not independent. Apr 26, 2016 at 1:48 • Part c is wrong, hence part d. It's good to make mistakes, and I'll point this one out. So part c says, first three coins are heads AND there are exactly five heads and tails. When you hear AND, you do not multiply by default. The events are dependent, because one depends on the other : if the first three tosses are heads, it doesn't allow the rest to the tosses to behave normally, because only 2 of the remaining 7 can be heads. Use this logic to refine c, and hence d, whose formula is correct. Apr 26, 2016 at 1:50 • Should I use the formula$P(B|A) = \frac{P(A \cap B)}{P(A)}$? Apr 26, 2016 at 1:50 • @dewick49 That would work. Apr 26, 2016 at 1:51 • Is there a simple way to find$P(A \cap B)$here? I'm having trouble finding it without writing the whole set out. Or is that essentially what Andre is saying in his answer? Apr 26, 2016 at 2:04 ## 1 Answer c) The probability the first$3$are heads is, as you wrote,$\frac{1}{8}$. The (conditional) probability that we end up with$5$of each, given that the first$3$were heads, is the probability of$2$heads in the next$7$throws. This is$\frac{\binom{7}{2}}{2^{7}}$. Thus our required probability is$\frac{1}{8}\cdot \frac{\binom{7}{2}}{2^{7}}$. Remark: a) and b) were right. For d), your method is fine, with suitable correction for$\Pr(A\cap B)$. • Is there a way to express this answer with the formula in the above comments, out of curiosity? Apr 26, 2016 at 2:09 • Yes, let$B$be the event first three are heads, and let$A$be the event$5$of each. We want$\Pr(A\cap B)$. Now use the fact that$\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}$, or equivalently$\Pr(A\cap B)=\Pr(A\mid B)\Pr(B)$. In my answer, I computed$\Pr(A\mid B)$as$\binom{7}{2}/2^7$, then multiplied by$1/8\$. So in essence I used the conditional probability formula, without thinking of what I was doing as using a formula. Apr 26, 2016 at 2:15
• Ah, I see! Fantastic. Thank you! Apr 26, 2016 at 2:18
• You are welcome. Apr 26, 2016 at 2:20