Question

Each of two persons $A$ and $B$ toss three fair coins. The probability that both get the same number of the head is
$\left( a \right){\text{ }}\dfrac{3}{8}$
$\left( b \right){\text{ }}\dfrac{1}{9}$
$\left( c \right){\text{ }}\dfrac{5}{{16}}$
$\left( d \right){\text{ }}\dfrac{7}{{16}}$

Answer 1

Hint:
For solving this type of question we need to first find the sample when three coins are tossed and since both are tossing so it will become the square of the sample. Then out of sample points in a toss, we will count each head pair and then square them, and on adding we will get the favorable outcome. And from this now we can easily calculate the probability.

Formula used:
Probability,
$P\left( A \right) = \dfrac{{Number{\text{ of favourable outcome}}}}{{Total{\text{ number of favourable outcomes}}}}$
Here,
$P(A)$ , will be the probability of any event named $A$

Complete step by step solution:
For tossing the $3$ coins the sample space will be
$ \Rightarrow S = \left\{ {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} \right\}$
Therefore, the total sample space will be of $n\left( s \right) = 8$
Since both $A\& B$ are tossing the coins, therefore, the sample space, in this case, will be
$ \Rightarrow n(s) = {8^2}$
And on solving, we get
$ \Rightarrow n(s) = 64$
Since, out of the $8$ sample points in a toss of the three coins, we have
$3$ have $2$ heads, $3$ have $1$ heads, $1$ has $2$ heads, and $1$ has $0$ heads.
Since we know that the tossing is performed by both $A\& B$ , therefore we will square the results and then add them.
$ \Rightarrow n(A) = {3^2} + {3^2} + {1^2} + {1^2}$
And on solving the above equation, we get
$ \Rightarrow n(A) = 20$
Now we will calculate the probability, and we already know he formula so on substituting the values, we get
$ \Rightarrow P\left( A \right) = \dfrac{{20}}{{64}}$
And solving for the simplest form further, we get
$P\left( A \right) = \dfrac{5}{{16}} \Rightarrow $

Hence, the option $\left( c \right)$ is correct.

Note:
Probability shows that the occasion is difficult to occur while having probability equivalent to $1$, shows that the occasion will surely occur. We should recall that the amount of probability of the event of some occasion and the probability of non-event of a similar occasion is consistent $1$. These are the basic things we should know about probability.