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Why is the probability of tossing a coin 10 times and getting 8 heads the same probability as tossing ten times and getting two heads
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It's a neat property of binomial probabilities.
it helps that P(H)=P(T)
Yup
P(8H)=P(8T)=P(2H)
In terms of the binomial formula: \[p_1=P(\text{8 heads in 10 tosses})=\binom{10}8\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^{2}\] Knowing that \(\dbinom{10}8=\dbinom{10}2\), you have \[p_2=P(\text{2 tails in 10 tosses})=\binom{10}2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{8}=p_1\]
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because the number of ways to include 2 out of 10 people is the same as the number of ways to exclude 8 out of 10
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