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A fair coin is tossed three times. What is the probability it will land on heads exactly twice?

We know that flipping a single coin has two mutually exclusive outcomes, and that multiple flips constitute independent events.

This language suffices to explain why, for the trivial case of two coin flips, we can compute the odds of landing heads twice as P(H and H) = 1/2 x 1/2 = 1/4.

What other language can we use to describe the primary question? What are some attributes of probability problems that can be solved using the binomial coefficient?

I'm hoping that reading said language will develop my intuition as to why we apply n-choose-k to solve this problem.

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    $\begingroup$ A few things... first, do not confuse the words "odds" and "probability." The odds of flipping two heads when flipping a coin twice is $1:3$. The probability of doing so is $\frac{1}{4}$. Next, for the problem in the title... this can be done trivially via a binomial distribution. The binomial coefficient appears in the result because of the number of different sequences of "heads" versus "tails" which all correspond to having flipped the desired number of heads. $\endgroup$ – JMoravitz Jun 8 '20 at 1:13
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    $\begingroup$ If you want to "develop your intuition" then try working out the "flipping four coins in sequence what is the probability of having flipped exactly two heads with an unfair coin who lands on heads with probability $p$" problem manually by making yourself a tree diagram, and recognizing the probability of two heads will be the sum of the probabilities: $Pr(HHTT)+Pr(HTHT)+Pr(HTTH)+Pr(THHT)+Pr(THTH)+Pr(TTHH)$, each of which can be broken down via independence and recognizing that each individual occurred with the same probability. Yes, this is tedious. That is why we simplify it. $\endgroup$ – JMoravitz Jun 8 '20 at 1:18
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The language is the language of 'random variables' and 'probability distributions'. These are fundamental ideas in probability theory.

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