0

If a fair coin is tossed 3 times, what is the probability that it turn up heads exactly twice?

Without having to list the coin like HHH, HHT, HTH, ect. to get to P=3/8. I would like to ask if there is any mathematical way to calculate this probability.

Please help, thank you!

Tui
  • 87
  • 1
    Listing all possibilities and then choose some among them for given event is very mathematical way. If you want name of formal model, then it is classical probability model on finite probability space. – zkutch Jun 26 '20 at 04:33
  • @zkutch thank you, but in an exam when time is not on our side, is there a formula that we can apply to? for example, "if a fair die is tossed 3 times, what is the probability that it turn up six exactly twice", we don't have enough time to list all of them. – Tui Jun 26 '20 at 04:36
  • 1
    In case, when you have only 8 possibilities I don't know which is more fast: use classical $\frac{|A|}{| \Omega |}$ or calculating some formula, which you need to recall. Up to person, may be. – zkutch Jun 26 '20 at 04:56

1 Answers1

2

Since it is a binomial experiment, the probability can be found quickly using $$P(k)={n\choose k}(1/2)^n$$ "if a fair die is tossed 3 times, what is the probability that it turn up six exactly twice" $$P(2)={3 \choose 2}(1/6)^2(5/6)^1.$$ The general formula is $$P(k)={n\choose k}p^k (1-p)^{n-k},$$ where $p$ is the probability of success.

toronto hrb
  • 880
  • Thank you! So the result for p^k * (1-p)^(n-k) = 5/216, not 15/216 because six can be like 66x, 6x6, x66? or (n k) is n, which is 3? – Tui Jun 26 '20 at 05:09
  • 1
    Yes. You have 3 positions and choose 2 for number 6. There are 3 cases as you listed: 66x, 6x6, and x66. Each of them has a probability of (1/6)(1/6)(5/6). Adding them together, you have $3(1/6)(1/6)(5/6)$. – toronto hrb Jun 26 '20 at 05:18
  • I don't understand, that's why I ask. I'm sorry if it bothered you. I read it here also https://math.stackexchange.com/a/1327209/748377, but then it doesn't make sense since the (3 3)(1/6)(1/6)(1/6) does not *3. – Tui Jun 26 '20 at 05:20
  • 1
    You can pick any statistics book and read its binomial distribution part. – toronto hrb Jun 26 '20 at 05:21
  • Ok, thank you!! – Tui Jun 26 '20 at 05:23
  • 1
    ${3\choose 3}(1/6)(1/6)(1/6)$ is the probability of turning up 6 for all 3 times, where ${3\choose 3}=1$. – toronto hrb Jun 26 '20 at 05:23
  • I'm sorry to bother you 1 last time, may I ask that when k = n then (n k) = 1, and when k < n then (n k) = n? – Tui Jun 26 '20 at 05:36