Question

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered $2,3,4,......12$ is picked and the number on the card is noted. The probability that the noted number is either $7$ and $8$ is $1k3/792$,where $k$ is ten's place of the number $1k3$. F ind the value of $k$. {$1k3$ =$100$+$10\times k$+$3$}

Answer 1

Let, $E_1$ be the event noted number is $7$.
$E_2$ be the event noted number is $8$.
$H$ be getting head on coin.
$T$ be getting tail on coin.
Therefore By law of total probability,
$P\left(E_1\right)=P\left(H\right).P\left(\frac{E_1}{H}\right)+P\left(T\right)P\left(\frac{E_1}{T}\right)$
and $P\left(E_2\right)=P\left(H\right).P\left(\frac{E_2}{H}\right)+P\left(T\right).P\left(\frac{E_2}{T}\right)$
where $P\left(H\right)=\frac{1}{2}=P\left(T\right)$
$P\left(\frac{E_1}{H}\right)=$ Probability of getting a sum of $7$ on two dice.
Here, favorable cases are $\{(1,6)(6,1),(2,5),(5,2),(3,4),(4,3)\}$
$\therefore P\left(\frac{E_1}{H}\right)=\frac{6}{36}=\frac{1}{6}$
Also, $P\left(\frac{E_1}{T}\right)=$ Probability of getting $7$ numbered card out of $11$ cards $=\frac{1}{11}$
$P\left(\frac{E_2}{H}\right)=$ Probability of getting a sum of $8$ on two dice, here favorable cases are $\{(2,6),(6,2),(4,4),(5,3),(3,5)\}$
$\therefore P\left(\frac{E_2}{H}\right)=\frac{5}{36}$
$P\left(\frac{E_2}{T}\right)=$ Probability of getting $8$ numbered card out of $11$ cards $=\frac{1}{11}$
$\therefore P\left(E_1\right)=(\frac{1}{2}\times \frac{1}{6})+(\frac{1}{2}\times \frac{1}{11})=\frac{1}{12}+\frac{1}{22}=\frac{17}{132}$
and $P\left(E_2\right)=(\frac{1}{2}\times \frac{5}{36})+(\frac{1}{2}\times \frac{1}{11})=\frac{1}{2}\left[\frac{91}{396}\right]=\frac{91}{729}$
Now $E_1$ and $E_2$ are mutually exclusive events.
Therefore, $P\left(E_{1}or{E}_2\right)=P\left(E_1\right)+P\left(E_2\right)=\frac{17}{132}+\frac{91}{792}=\frac{193}{792}$