A coin is tossed 100 times. What is the proba | vedclass

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(A) Let $p$ be the probability of getting a tail,so $p = \frac{1}{2}$.
Let $q$ be the probability of getting a head,so $q = \frac{1}{2}$.
Here,$n =

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$\frac{1}{2}$

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(A) Let $p$ be the probability of getting a tail,so $p = \frac{1}{2}$.
Let $q$ be the probability of getting a head,so $q = \frac{1}{2}$.
Here,$n = 100$ trials are performed.
The probability of getting a tail $k$ times is given by the binomial distribution: $P(X = k) = ^{100}C_k p^k q^{100-k}$.
We need the probability of getting an odd number of tails,which is $P(X = 1) + P(X = 3) + \dots + P(X = 99)$.
Using the binomial expansion, we know that $(q + p)^n = \sum_{k=0}^{n} {^{n}C_k} p^k q^{n-k}$ and $(q - p)^n = \sum_{k=0}^{n} {^{n}C_k} (-p)^k q^{n-k}$.
Subtracting these two equations: $(q + p)^n - (q - p)^n = 2 	imes [{^{n}C_1} p^1 q^{n-1} + {^{n}C_3} p^3 q^{n-3} + \dots]$.
Since $p = q = \frac{1}{2}$,we have $q + p = 1$ and $q - p = 0$.
Therefore,the sum of probabilities for odd $k$ is $\frac{(q + p)^n - (q - p)^n}{2} = \frac{1^n - 0^n}{2} = \frac{1}{2}$.

$A$ coin is tossed $100$ times. What is the probability of getting a tail an odd number of times?

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