those the probability of acquiring 3 heads and also 7 tails if one flips a same coin 10 times. I just can’t number out how to version this correctly.

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Your concern is related to the binomial distribution.

You do $n = 10$ trials. The probability of one successful trial is $p = frac12$. You want $k = 3$ successes and also $n - k = 7$ failures. The probability is:

$$inomnk p^k (1-p)^n-k = inom103 cdot left(dfrac12 ight)^3 cdot left(dfrac12 ight)^7 = dfrac15128$$

One method to know this formula: You want $k$ successes (probability: $p^k$) and also $n-k$ failure (probability: $(1-p)^n-k$). The successes have the right to occur almost everywhere in the trials, and also there space $inomnk$ to arrange $k$ successes in $n$ trials.

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edited might 30 "12 in ~ 22:51
answered might 30 "12 in ~ 22:45

Ayman HouriehAyman Hourieh
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We develop a chrischona2015.orgematical design of the experiment. Compose H because that head and T because that tail. Document the results of the tosses as a string of length $10$, consisted of of the letters H and/or T. So for instance the cable HHHTTHHTHT means that we obtained a head, then a head, then a head, then a tail, and so on.

There are $2^10$ together strings of length $10$. This is due to the fact that we have $2$ choices for the first letter, and also for every together choice we have actually $2$ choices for the 2nd letter, and for every an option of the an initial two letters, we have actually $2$ choices for the third letter, and also so on.

Because us assume the the coin is fair, and that the an outcome we gain on say the first $6$ tosses go not affect the probability of acquiring a head ~ above the $7$-th toss, every of these $2^10$ ($1024$) strings is equally likely. Because the probabilities must include up to $1$, every string has probability $frac12^10$.So for instance the outcome HHHHHHHHHH is just as likely as the result HTTHHTHTHT. This may have an intuition implausible feel, however it fits in really well with experiments.

Now let united state assume the we will be happy just if we get specifically $3$ heads. To uncover the probability we will certainly be happy, we count the number of strings that will certainly make united state happy. Suppose there are $k$ together strings. Climate the probability we will be happy is $frack2^10$.

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Now we require to discover $k$. So we should count the variety of strings that have precisely $3$ H"s. To carry out this, we uncover the variety of ways come choose where the H"s will occur. So us must pick $3$ locations (from the $10$ available) for the H"s come be.

We can pick $3$ objects indigenous $10$ in $inom103$ ways. This number is called likewise by various other names, such together $C_3^10$, or $_10C_3$, or $C(10,3)$, and also there are various other names too. That is called a binomial coefficient, since it is the coefficient of $x^3$ once the expression $(1+x)^10$ is expanded.

There is a useful formula for the binomial coefficients. In general$$inomnr=fracn!r!(n-r)!.$$

In particular, $inom103=frac10!3!7!$. This turns out to be $120$. Therefore the probability of specifically $3$ top in $10$ tosses is$frac1201024$.

Remark: The idea deserve to be dramatically generalized. If us toss a coin $n$ times, and also the probability of a head on any kind of toss is $p$ (which require not be same to $1/2$, the coin could be unfair), climate the probability of precisely $k$ top is$$inomnkp^k(1-p)^n-k.$$This probability version is dubbed the Binomial distribution. That is of an excellent practical importance, since it underlies all an easy yes/no polling.