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

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.

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