Mathematics - Probability Question with Solution | TestHub

Let, $A$ be the event that the noted number is either $7$ or $8.$

And, $E_1$ be the event that the toss of coin results in head and $E_2$ be the event that the toss of coin results in tail.

Then, we have $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$

And, $P\left(A|E_1\right)=$$P($getting the sum of the numbers on the pair of dice as $7$ or $8)$

The total number of cases, when a pair of dice is thrown are $=6\times 6=36$ and the cases of getting sum $7$ or $8$ are $\left\{\left(1, 6\right), \left(2, 5\right), \left(3, 4\right), \left(4, 3\right), \left(5, 2\right), \left(6, 1\right), \left(2, 6\right), \left(3, 5\right), \left(4, 4\right), \left(5, 3\right), \left(6, 2\right)\right\}=11$ cases.

$\Rightarrow P\left(A|E_1\right)=\frac{11}{36}$

Similarly, $P\left(A|E_2\right)=($getting the number $7$ or $8$ from the numbers $1, 2, 3,.., 9)$

$\Rightarrow P\left(A|E_2\right)=\frac{2}{9}$

Now, using the total probability theorem, we get

$P\left(A\right)=P\left(E_1\right)P\left(A|E_1\right)+P\left(E_1\right)P\left(A|E_2\right)$

$\Rightarrow P\left(A\right)=\frac{1}{2}\times \frac{11}{36}+\frac{1}{2}\times \frac{2}{9}$

$\Rightarrow P\left(A\right)=\frac{11}{72}+\frac{1}{9}=\frac{19}{72}.$