Explanation of task from the probability quiz

Hello, I studied this quiz.


Which of these events is the most likely to happen when flipping a fair coin?

Flip 2 or more heads when flipping 3 coins.

Flip 20 or more heads when flipping 30 coins.

Flip 200 or more heads when flipping 300 coins.

Instead of making an explicit calculation, think about how likely or unlikely each of these outcomes would be.

Explanation Correct answer: 2 out of 3

When you flip some number of coins, you expect about half of them to be heads. As the number of coins increases, you should expect the observed proportion of heads to be closer and closer to 50\%.50%. The casual logic is: sure, weird things can happen a few times, but not hundreds of times.

With more probability experience, you can formalize and quantify the ideas behind this problem.

Here are the actual probabilities for completeness:

P(\text{2 or more heads out of 3})=0.5P(2 or more heads out of 3)=0.5

P(\text{20 or more heads out of 30})\approx 0.05P(20 or more heads out of 30)≈0.05

P(\text{200 or more heads out of 300})\approx 0.000000004P(200 or more heads out of 300)≈0.000000004


And i can't find the explanation , how were those probabilities calculated? In this explanation i can only see formula and result

Note by Paul Gavrilov
1 month, 1 week ago

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1 vote

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See the binomial distribution.

Example for 33 flips:

If NN is the number of heads after 3\color{#3D99F6}3 flips, then NB(3,0.5)N\sim B(\color{#3D99F6}3\color{#333333}, 0.5). Using the cumulative distribution (on your calculator, for 30 or 300 flips), Pr(N2)=Pr(N=2)+Pr(N=3)=(32)(0.5)2(10.5)32+(33)(0.5)3(10.5)13=38+18=12\text{Pr}(N\geqslant 2)=\text{Pr}(N=2)+\text{Pr}(N=3)=\binom{3}{2}(0.5)^{2}(1-0.5)^{3-2}+\binom{3}{3}(0.5)^3(1-0.5)^{1-3}=\dfrac{3}{8}+\dfrac{1}{8}=\color{#20A900}\boxed{\dfrac{1}{2}}.

Alternatively, Pr(n or more heads from k flips)=12kr=nk(kr)\text{Pr(n or more heads from k flips)}=\frac{1}{2^k}\sum_{r=n}^{k}\binom{k}{r}

As there are 2k2^k possible and equally likely outcomes (2 for the first flip, 2 for the second flip, ..., 2 for the kkth flip) and (kr)\binom{k}{r} ways to get rr coins.

Matthew Christopher - 1 month, 1 week ago

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I know.

I get this question wrong as well!

Yajat Shamji - 1 month, 1 week ago

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