Suppose the probability that a coin toss shows “head” is p, where 0<p<1. The coin is tossed repeatedly until the first “head” appears. The expected number of tosses required is

The correct option is 1/p.

To find the expected number of tosses required to get the first head, we can model this process using a geometric distribution. Let $X$ be the random variable representing the number of tosses required to obtain the first "head".

The probability of getting a head on any individual toss is given as $p$, and the probability of getting a tail is $1-p$. The first head occurs on the $k$-th toss if and only if we get $k-1$ consecutive tails followed by a head. Therefore, the probability mass function of $X$ is:

$$P(X=k)={(1-p)}^{k-1}p$$

where $k=1,2,3,...$

The expected value $E\left[X\right]$ is the sum of the product of each possible outcome and its probability:

$$E\left[X\right]=\sum _{k=1}^{\infty }k·P(X=k)=\sum _{k=1}^{\infty }k{(1-p)}^{k-1}p$$

We can factor out the constant $p$ from the summation:

$$E\left[X\right]=p\sum _{k=1}^{\infty }k{(1-p)}^{k-1}$$

To evaluate the infinite series, let $q=1-p$. Since $0<p<1$, we have $0<q<1$. The series becomes $\sum _{k=1}^{\infty }k{q}^{k-1}$.
We know the sum of a standard infinite geometric series is:

$$\sum _{k=0}^{\infty }{q}^k=\frac{1}{1-q}$$

Differentiating both sides with respect to $q$, we obtain:

$$\sum _{k=1}^{\infty }k{q}^{k-1}=\frac{d}{dq}{(1-q)}^{-1}=\frac{1}{{(1-q)}^2}$$

Substituting $1-q=p$ back into this equation gives:

$$\sum _{k=1}^{\infty }k{(1-p)}^{k-1}=\frac{1}{p^2}$$

Finally, we plug this result back into our expectation formula:

$$E\left[X\right]=p·\frac{1}{p^2}=\frac{1}{p}$$

Thus, the expected number of tosses required to get the first head is $\frac{1}{p}$.