Question

A coin is tossed successively until the ${1^{st}}$ time head occurs. The expected number of tosses required is
A. $4$
B. $2$
C. $1$
D. $5$

Answer 1

As the coin is tossed successively until for the ${1^{st}}$ time head occurs, this is the case of expected wait. We solve this question by using the expected value of all the possibilities of the tosses.

Complete step-by-step answer:
In the question, it is given that A coin is tossed successively until the ${1^{st}}$ time head occurs, so we will use the probability function in this.
Let $X$ be the number of tosses of a fair coin until a head appears, and we want to find the expected value of $X$ i.e., $E(X)$ . When we toss the coin once, there are two possibilities:
Possibility $1$: The first toss is heads: In this case, the value of $X$ will be $1$.
Possibility $2$: The first toss is tails: In this case, we have lost one trial, but since we needed a head in the first trial, we are back to where we started from. So, the expected number of trials until heads come in the first trial will be equal to $1$ (from the lost trial) plus $E(X)$ .
Therefore, probability of getting a head in a toss ${P_H} = \dfrac{1}{2}$ , as there are a total of only two choices- we can either get a head or a tail.
And, the expected value is the sum of: (each of the possible outcomes) $\times$ (the probability of the outcome occurring)
$E(X) = \dfrac{1}{2}(1) + \dfrac{1}{2}(1 + E(X))$
Or, $E(X) - \dfrac{1}{2}.E(X) = 1$
Or, $\dfrac{1}{2}.E(X) = 1$
Or, $E(X) = 2$

So, the correct answer is “Option B”.

Note: The expected value (or mean) of $X$ , where $X$ is a discrete random variable, is a weighted average of the possible values that $X$ can take, each value being weighted according to the probability of that event occurring. The expected value of $X$ is usually written as $E(X)$ or $m$.
$E\left( X \right) = Sx.P\left( {X = {\text{ }}x} \right)$
So, the expected value is the sum of: (each of the possible outcomes) $\times$ (the probability of the outcome occurring).

Alternate Solution:
Either the event occurs on the first trial with probability $p$ , or with probability $1 - p$ the expected wait is $1 + E$
$E = p + (1 - p)(1 + E)$
Or, $E = \dfrac{1}{p}$
Therefore, probability of getting a head in a toss ${P_H} = \dfrac{1}{2}$ , as there are a total of only two choices- we can either get a head or a tail.
So, $E = 2$ .
Therefore, (B.) $2$ is the required answer.