Flipping A Coin 10 Times

Probability Level 2

Azalea flips a fair coin 10 times. What is the probability that she gets heads in at least 8 of the 10 flips?

9/127 7/128 3/31 33/73

1 solution

Hayden Park
Jun 5, 2020

There are $2^{10} = 1024$ possible outcomes of the 10 coin flips. The number of ways to get 8, 9, or 10 heads is $\binom{10}{8}+\binom{10}{9}+\binom{10}{10}=45+10+1=56$ . So the probability that Azalea gets heads at least 8 times is $\dfrac{56}{1024}=\boxed{\dfrac{7}{128}}$ .

Which part of my solution is wrong. I know it is wrong because it is not so likely

Total possibilities = $2^{10} = 1024$ . Getting at least $\binom{10}{8}$ selection spots $\times 2 \times 2$ because the two spots left can have any of heads or tails

Thus probability $= \boxed{\dfrac{45}{256}}$

Mahdi Raza - 1 year ago

I believe you are counting certain combinations multiple times. If the problem were 4 coin flips, what the probability of at least 3 heads the answer is $\displaystyle \frac{5}{16}$ .

$\begin{matrix} H & H & H & H \\ H & H & H & T \\ H & H & T & H \\ H & T & H & H \\ T & H & H & H \\ \end{matrix}$

Your solution implies find the number of ways of selecting 3 heads out of 4 flips $\displaystyle \binom{4}{3}$

$\begin{matrix} H & H & H & \Box \\ H & H & \Box & H \\ H & \Box & H & H \\ \Box & H & H & H \\ \end{matrix}$

Then 2 Choices for each of the unchosen boxes $\Box$ gives the following:

$\begin{matrix} H & H & H & \mathbf{T} \\ H & H & H & \mathbf{H} \\ H & H & \mathbf{T} & H \\ H & H & \mathbf{H} & H \\ H & \mathbf{T} & H & H \\ H & \mathbf{H} & H & H \\ \mathbf{T} & H & H & H \\ \mathbf{H} & H & H & H \\ \end{matrix}$

Its now clear 4 heads have been counted 3 extra times. Hope that helps.

Eric Roberts - 12 months ago

Thanks! My idea counted "All heads" cases many times. That helped

Mahdi Raza - 12 months ago

@Mahdi Raza – Hi Mahdi Raza! It also would have overcounted counted many other combinations as well, not just heads. There are 56 ways in total, you would have produced 180 combinations in total ( 4 ways for each of the 45 ways of selecting 8 out of 10 ). However, The "All heads" combination can only be produced 45 times ( 1 out of the 4 ways for each of the 45 ways of selecting 8 of 10 of which you would toss out 44 extras). "All heads" doesn't account for the disparity between the 136 remaining combinations ( after 44 of them are removed ) and the 56 you need to pair down to. From your reply I wasn't sure if my "workable" example was misleading you, so I thought I should clarify.

Eric Roberts - 12 months ago

@Eric Roberts – Hi Eric, It wasn't misleading, I knew that there were other double-counted cases as well along with all heads, I just didn't say that in my comment. But, thanks for clarifying!!

Mahdi Raza - 12 months ago

Flipping A Coin 10 Times

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