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 the geometric distribution. Let $X$ be the random variable representing the number of tosses required until the first "head" appears.

The probability that a coin toss shows "head" is given as $p$ (where $0<p<1$). Consequently, the probability of getting a "tail" on any single toss is $1-p$.

For the first head to appear on the $k$-th toss, the first $k-1$ tosses must result in tails, and the $k$-th toss must result in a head. The probability of this sequence is:
$P(X=k)={(1-p)}^{k-1}p$
where $k=1,2,3,...$

The expected value $E\left[X\right]$ is the weighted average of all possible outcomes:
$E\left[X\right]=\sum _{k=1}^{\infty }k\cdot P(X=k)$
$E\left[X\right]=\sum _{k=1}^{\infty }k\cdot {(1-p)}^{k-1}p$

We can factor out the constant $p$ from the summation:
$E\left[X\right]=p\sum _{k=1}^{\infty }k\cdot {(1-p)}^{k-1}$

Let $q=1-p$. The sum is then:
$S=\sum _{k=1}^{\infty }k\cdot q^{k-1}=1+2q+3q^2+4q^3+...$

This is an arithmetico-geometric series. We can evaluate it by multiplying $S$ by $q$:
$qS=q+2q^2+3q^3+...$

Subtracting $qS$ from $S$:
$S-qS=1+q+q^2+q^3+...$
$S(1-q)=\frac{1}{1-q}$ (since $0<q<1$)

Thus, we find:
$S=\frac{1}{{(1-q)}^2}$

Substituting $q=1-p$ back, we get:
$S=\frac{1}{p^2}$

Now substitute this back into the expectation formula:
$E\left[X\right]=p\cdot S=p\cdot \frac{1}{p^2}=\frac{1}{p}$

Therefore, the expected number of tosses required is indeed 1/p.