what’s the probability of obtaining 3 heads and 7 tails if one flips a fair 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 the one effective trial is $p = frac12$. You desire $k = 3$ 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 recognize this formula: You want $k$ successes (probability: $p^k$) and also $n-k$ failure (probability: $(1-p)^n-k$). The successes can occur almost everywhere in the trials, and also there are $inomnk$ come arrange $k$ successes in $n$ trials.

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

Ayman HouriehAyman Hourieh
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We develop a discoverhotmail.comematical version of the experiment. Compose H because that head and also T for tail. Document the results of the tosses as a wire of size $10$, made up of the letter H and/or T. Therefore for instance the wire HHHTTHHTHT method that we gained a head, climate a head, then a head, climate a tail, and also so on.

There are $2^10$ together strings of size $10$. This is due to the fact that we have actually $2$ options for the an initial letter, and for every such choice we have actually $2$ options for the second letter, and also for every choice of the first two letters, we have actually $2$ options for the 3rd letter, and so on.

Because we assume that the coin is fair, and also that the result we get on say the an initial $6$ tosses does not impact the probability of gaining a head top top the $7$-th toss, each of this $2^10$ ($1024$) strings is equally likely. Due to the fact that the probabilities must include up come $1$, every string has actually probability $frac12^10$.So for instance the outcome HHHHHHHHHH is just as most likely as the result HTTHHTHTHT. This may have actually an intuition implausible feel, however it fits in very well through experiments.

Now let us assume the we will certainly be happy only if us get precisely $3$ heads. To find the probability we will be happy, us count the number of strings that will make united state happy. Suppose there space $k$ together strings. Climate the probability we will certainly be happy is $frack2^10$.

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Now we require to discover $k$. For this reason we must count the variety of strings the have precisely $3$ H"s. To execute this, we discover the variety of ways come choose where the H"s will occur. So us must select $3$ places (from the $10$ available) for the H"s come be.

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

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

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

Remark: The idea can be considerably generalized. If we toss a coin $n$ times, and 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), then the probability of exactly $k$ top is$$inomnkp^k(1-p)^n-k.$$This probability model is called the Binomial distribution. That is of good practical importance, because it underlies all simple yes/no polling.