A fair coin is tossed 100 times. What is the probability that more than 55 heads are observed?

A fair coin is tossed 100 times. What is the probability that more than 55 heads are observed?

I need a clarification on how to use binomial distribution formula in this problem. What i got was:

$$x = 55, \quad \mu = 50, \quad SD = 5$$ $$z = (55-50)/5 \implies z = 0.84 \implies 1- z = 0.16$$ $$p(x) = 0.16$$

The exact answer is obtained from the Binomial distribution. Various pieces of software, including Wolfram Alpha, will do the computation for you. This approach is painful to carry out by hand.

Perhaps you are expected to use the normal approximation to the binomial. If $X$ is the number of heads, then $X$ has approximately normal distribution, mean $50$, standard deviation $\sqrt{100(1/2)(1/2)}=5$.

We want to find the probability that $X\ge 56$. We find, approximately, the probability of the complement, the probability that $X\le 55$. This, approximately, is the probability that a normal with mean $50$ and standard deviation $5$ is $\le 55$.

But perhaps you are expected to use the continuity correction. Then $$\Pr(X\le 55)\approx\Pr\left(Z\le \frac{55.5-50}{5}\right)\tag{1}$$ where $Z$ is standard normal. If you are not expected to use the continuity correction, use $55$ instead of $55.5$.

In either case, the answer to the original question is, approximately, $1$ minus the number obtained in (1).

• so what i got for Less than and equal to 55 is 0.84, for 56 above would be 1-0.84= 0.16? to obtain more than 55 heads? – Kit lai Aug 16 '14 at 4:15
• Yes, if you do not use the continuity correction, it is $\approx 1-0.8413$. If you use the correction, it is $\approx 1-0.8643$. Note how close this gets us to the exact answer. – André Nicolas Aug 16 '14 at 4:18
• thanks , i Kinda got what you were you trying to say – Kit lai Aug 16 '14 at 4:27
• You are welcome. – André Nicolas Aug 16 '14 at 4:27
• @AndréNicolas I just want make sure. We can use Chernoff Bound instead and we get $1/e^{1/6}$ – YOUSEFY May 25 '17 at 11:26

As lab bhattacharjee mentioned, the probability of exactly $k$ heads is

$\binom{n}{k} p^k (1-p)^{n-k}$

Let us write $X$ to denote the number of heads, and $Q$ to denote the desired probability, $P(X > 55)$. Note that by symmetry we have $P(X > 55) = P(X < 45)$. Also,

$P(X > 55) + P(X < 45) + P(45 \leq X \leq 55) = 1$

so $2Q + P(45 \leq X \leq 55) = 1$. Therefore the problem reduces to evaluating $Z = P(45 \leq X \leq 55)$.

We can use another symmetry argument to evaluate $Z$ by observing that $P(45 \leq X < 50) = P(50 < X \leq 55)$. This means that we will only need to calculate $\binom{n}{k} p^k (1-p)^{n-k}$ for $50 \leq k \leq 55$, substantially reducing the workload.

• so p is the probability for obtaining H or T which is 1/2? n = 100 trials, k = 55 heads ? – Kit lai Aug 16 '14 at 4:33
• Yes, that's right, p = 1/2, n = 100. k = number of heads. – Bungo Aug 16 '14 at 4:44

HINT:

From this,

the probability of getting exactly $k$ successes in $n$ trials is $$\binom nk p^k(1-p)^{n-k}$$

Here $\displaystyle p=\dfrac12\implies p^k(1-p)^{n-k}=\left(\frac12\right)^n, n=100$

and $\displaystyle\sum_{k>55}^{100}\binom nk=\frac{(1+1)^{100}}2-\sum_{k=50}^{55}\binom nk$

• my answer was 0.0048 – Kit lai Aug 16 '14 at 3:46
• ^That's a bit off. See this – JimmyK4542 Aug 16 '14 at 3:48
• @JimmyK4542 i dont really get why was (Sum[Binomial[100, k]/2^100, {k, 56, 100}]) 2^100 where did the 2 come from or from which formula? – Kit lai Aug 16 '14 at 3:58
• @Kitlai, $$\sum_{k=0}^n\binom nk=(1+1)^n$$ and $$\binom nk=\binom n{n-k}\implies\sum_{k=0}^{49}\binom nk=\sum_{k=51}^{100}\binom nk$$ – lab bhattacharjee Aug 16 '14 at 4:01