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Problem Statement: You have 21 fair coins, each flipped once. What is the probability of obtaining an even number of heads?
Understanding the Problem:
Key Insight: The probability of getting an even number of heads is equal to the probability of getting an odd number of heads. This is due to the symmetric nature of the binomial distribution when the probability of success (getting a head) is 0.5.
Mathematical Explanation:
Binomial Distribution:
The number of heads in 21 flips follows a binomial distribution with parameters n=21 and p=0.5.
The probability of getting exactly k heads is given by:
P(k)=(k21)⋅(0.5)k⋅(0.5)21−k=(k21)⋅(0.5)21
Even vs Odd Heads:
We are interested in the sum of probabilities for even k: 0, 2, 4, ..., 20.
By symmetry, the sum of probabilities for odd k: 1, 3, 5, ..., 21 is the same as for even k.
The total probability for all possible outcomes (even and odd) is 1, so:
P(even)+P(odd)=1
Since P(even)=P(odd), it follows that:
2⋅P(even)=1⟹P(even)=0.5
Verification through Binomial Expansion:
Using the binomial theorem, the expansion of (1+x)21 gives:
(1+x)21=∑k=021(k21)xk
Setting x=1 gives the sum of all probabilities (equal to 2^21).
Setting x=−1 and adding gives the sum of coefficients for even k:
(1+1)21+(1−1)21=221+0=2⋅∑even k(k21)
Dividing by 2 gives the sum of probabilities for even k, confirming P(even)=0.5.
Conclusion: