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Hint: In this question first find out the probability of getting a head as well as not getting a head when a coin is tossed, later on in the solution construct all possible cases for getting a head as well as not getting a head because this time coin is tossed two times, so use this concept to reach the solution of the question.

Complete step-by-step answer:

Given a coin is tossed twice.

As we know a coin has two sides.

So, total number of outcomes $ = 2$

To get a head when a coin is tossed (p)$ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{1}{2}$

For not getting a head when a coin is tossed $\left( q \right) = 1 - \left( p \right)$

As the total probability is always 1.

$ \Rightarrow \left( q \right) = 1 - \left( p \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$

As the coin is tossed twice.

So, construct different cases for getting a head.

$1.$For not getting any head i.e. when we tossed the coin twice head does not appears$ = q.q = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}$

$2.$Head appears only once, it can happen in two ways when we tossed the first time we got head and in the second time head does not appear, and when we tossed again the first time we did not got head but in second time we got a head.

$ = pq + qp = \dfrac{1}{2}.\dfrac{1}{2} + \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} = \dfrac{1}{2}$

$3.$Always get a head i.e. when we tossed the coin twice head appears all the time $ = p.p = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}$

So, the required probability distribution is shown below.

$\begin{array}{*{20}{c}}

{\left( H \right)}&0&1&2 \\

{\left( {{P_h}} \right)}&{\dfrac{1}{4}}&{\dfrac{1}{2}}&{\dfrac{1}{4}}

\end{array}$

So, this is the required probability distribution of the number of heads (H).

Note: In such types of questions the key concept we have to remember is that always recall the property of probability which is stated above, then construct different cases as above to get the required probability distribution of the number of heads as shown above which is the required answer.

Complete step-by-step answer:

Given a coin is tossed twice.

As we know a coin has two sides.

So, total number of outcomes $ = 2$

To get a head when a coin is tossed (p)$ = \dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{1}{2}$

For not getting a head when a coin is tossed $\left( q \right) = 1 - \left( p \right)$

As the total probability is always 1.

$ \Rightarrow \left( q \right) = 1 - \left( p \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$

As the coin is tossed twice.

So, construct different cases for getting a head.

$1.$For not getting any head i.e. when we tossed the coin twice head does not appears$ = q.q = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}$

$2.$Head appears only once, it can happen in two ways when we tossed the first time we got head and in the second time head does not appear, and when we tossed again the first time we did not got head but in second time we got a head.

$ = pq + qp = \dfrac{1}{2}.\dfrac{1}{2} + \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} = \dfrac{1}{2}$

$3.$Always get a head i.e. when we tossed the coin twice head appears all the time $ = p.p = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}$

So, the required probability distribution is shown below.

$\begin{array}{*{20}{c}}

{\left( H \right)}&0&1&2 \\

{\left( {{P_h}} \right)}&{\dfrac{1}{4}}&{\dfrac{1}{2}}&{\dfrac{1}{4}}

\end{array}$

So, this is the required probability distribution of the number of heads (H).

Note: In such types of questions the key concept we have to remember is that always recall the property of probability which is stated above, then construct different cases as above to get the required probability distribution of the number of heads as shown above which is the required answer.

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