# Probability of 3 heads in 4 coin flips

Forgive me English is not my first language so this is confusing?

Suppose you toss a coin 4 times.
a. What is the probability that you will get exactly 3 heads?
b. Explain what P*(not 3 heads) means

*P meaning Probability

Can anyone explain how I would solve this word problem? Thank you!!

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Hint: What is the probability that you will get exactly $0$ heads? What is the probability that you will get exactly one head? If it helps, there are $2^4$ possibilities for the sequence of four flips. Try writing them all out and see if you can spot a pattern.

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(a) There are four different outcomes that give you exactly $3$ heads: THHH, HTHH, HHTH, and HHHT. The probability of getting exactly $3$ heads is the probability of getting one of these four outcomes, so it is

$$\frac4{\text{number of possible outcomes}};$$

what is the total number of possible outcomes?

(b) You’re tossing four coins, so if you don’t get $3$ heads, what are the possible numbers of heads that you could get? $P(\text{not }3\text{ heads})$ is the probability of getting one of these other numbers of heads.

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Thank you! The total number of possible outcomes would be 16, so it would be 4/16 or 1/4? – Miki Dec 13 '12 at 2:48
@Miki: That’s exactly right. – Brian M. Scott Dec 13 '12 at 2:49
Thank you very much, I really appreciate it!! (: – Miki Dec 13 '12 at 2:50
@Miki: You’re very welcome. – Brian M. Scott Dec 13 '12 at 3:15
@radicalmatt: Yes. For the first problem, for example, there are $2^4$ possible outcomes, all equally likely. There are $\binom43$ ways to pick $3$ of the $4$ tosses to be heads, so the probability if $\binom43/2^4=4/16=1/4$. – Brian M. Scott Jul 8 '15 at 19:16

well by calculating, using the formula $\frac{n!}{s!t!}\times(\frac{1}{2})^s \times (\frac{1}{2})^t$ where $n=$number of tosses s=probability for head $t=$probability for tail one gets answer to be $\frac{1}{4}$. (pretty cool huh)

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