Calculating the probability that more of a fair coin tosses will result in heads than in tails.

Question

My thought was the probability is: $\frac{5}{8} + \frac{6}{8} + \frac{7}{8} + 1$ and the sum is $\frac{13}{4}$ but this is not one of the choices. and the correct answer is E. I do not know why my thought is wrong and why the right answer is E, could anyone explain this for me please?

Answer 1

Hint 1: $$P(\text{more heads than tails}) + P(\text{more tails than heads}) + P(\text{same number of heads and tails}) = 1$$

Hint 2: $$P(\text{more heads than tails}) = P(\text{more tails than heads})$$

Hint 3: $$P(\text{same number of heads and tails}) = \binom{8}{4}\frac{1}{2^8}$$ (by the binomial distribution).

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Alternatively, the answer can be computed as $$P(\text{5 heads}) + P(\text{6 heads}) + P(\text{7 heads}) + P(\text{8 heads}) =\binom{8}{5} \frac{1}{2^8} + \binom{8}{6} \frac{1}{2^8} + \binom{8}{7} \frac{1}{2^8} + \binom{8}{8} \frac{1}{2^8}$$

Answer 2

Your reasoning is incorrect because of one thing: combinations (also, \frac{13}{4} is bigger than $1$ and certainly not a probability).

We need to count the number of ways each coin toss can happen. Recall that the number of ways to select $k$ objects from $n$ is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. This counts, in this case, the number of ways we can have, say, five heads in eight, is $\binom{8}{5} = 56.$ Calculations for six, seven, and eight heads in eight are done similarly.

So how do we finish this out? Easy,

$$ P(H > T) = \sum_{j=5}^8 \binom{8}{j} \left( \frac{1}{2} \right)^j \left( \frac{1}{2} \right)^{8-j} $$

But let's be a bit clever and instead say

$$ P(H > T) = \sum_{j=5}^8 \binom{8}{j} \left( \frac{1}{2} \right)^8 $$

Can you finish it from here?

Answer 3

There is a $\frac {70}{256}$ chance of getting exactly 4 heads.

It is equally likely to get more than 4 head as less than 4 heads.

$\frac {128-35}{256} = \frac {93}{256}$