We flip a fair coin 10 times. What is the probability that we get heads in exactly 8 of the 10 flips?

Question

We flip a fair coin 10 times. What is the probability that we get heads in exactly 8 of the 10 flips?

I thought the answer was this: P (Heads ≥ 8 flips) = P( Tails ≤ 2 flips ) =

C(10,0)(1/2)^10 + C(10,1) (1/2)^10 + C(10,2)(1/2)^10 = 7 / 128

But it is wrong. Can someone help?

Thanks in advance

Answer 1

probability of getting head tossing the coin is 1/2 and also probability of getting head tossing the coin is 1/2

and same sample space for n coins tossed 1 time or 1 coin tossed n times.

the probability of exactly 8 heads is

click on this image to the step-wise solution

Answer 2

You need to use binomial theorem to solve this problem.

$n$ = number of tosses (here, $n = 10$)

$p$ = probability of success (here, $p = {1 \over 2}$ as you have a fair coin)

$q$ = probability of failure (here, $q = 1 - p = {1 \over 2}$ )

The probability of exactly $k$ heads in $n$ tosses with a fair coin is given by

$ P(X = k) = \left( \matrix{ n \cr k \cr} \right) \ p^k q^{n - k}$

Taking $k = 8$ (exactly 8 heads), we get the required probability as follows:

$P(X = 8) = \left( \matrix{ 10 \cr 8 \cr} \right) \ \left( {1 \over 2} \right)^8 \left( {1 \over 2} \right)^{(10 - 8)} $

Note that

$\left( \matrix{ 10 \cr 8 \cr} \right) = \left( \matrix{ 10 \cr 2 \cr} \right) = {10 \times 9 \over 2 \times 1} = 45$.

Thus,

$P(X = 8) = 45 \left( {1 \over 2} \right)^8 \left( {1 \over 2} \right)^{2} $

i.e.

$P(X = 8) = 45 \left( {1 \over 2} \right)^{10}$

Simplifying, we get

$P(X = 8) = {45 \over 2^{10}} = {45 \over 1024} = 0.0439$.