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

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.

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.