JEE Main 2019 — Mathematics Probability & Statistics

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

And, ${E}_{1}$E1​ be the event that the toss of coin results in head and ${E}_{2}$E2​ be the event that the toss of coin results in tail.

Then, we have $P({E}_{1})=P({E}_{2})=\frac{1}{2}$P(E1​)=P(E2​)=21​

And, $P(A|{E}_{1})=$P(A∣E1​)=$P($P(getting the sum of the numbers on the pair of dice as $7$7 or $8)$8)

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

$$\Rightarrow P(A|{E}_{1})=\frac{11}{36}$$⇒P(A∣E1​)=3611​

Similarly, $P(A|{E}_{2})=($P(A∣E2​)=(getting the number $7$7 or $8$8 from the numbers $1,2,3,..,9)$1,2,3,..,9)

$$\Rightarrow P(A|{E}_{2})=\frac{2}{9}$$⇒P(A∣E2​)=92​

Now, using the total probability theorem, we get

$$P(A)=P({E}_{1})P(A|{E}_{1})+P({E}_{1})P(A|{E}_{2})$$P(A)=P(E1​)P(A∣E1​)+P(E1​)P(A∣E2​)

$$\Rightarrow P(A)=\frac{1}{2}\times \frac{11}{36}+\frac{1}{2}\times \frac{2}{9}$$⇒P(A)=21​×3611​+21​×92​

$$\Rightarrow P(A)=\frac{11}{72}+\frac{1}{9}=\frac{19}{72}.$$⇒P(A)=7211​+91​=7219​.