Probability Ques 78
An unbiased coin is tossed. If the result in 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 $2,3,4, \ldots, 12$ is picked and the number on the card is noted. What is the probability that the noted number is either $7$ or $8$ ?
$(1994,5$ M)
Show Answer
Answer:
Correct Answer: 78.$(\frac{193}{792})$
Solution:
Formula:
- Let, $E _1=$ the event noted number is $7$
$E _2=$ the event noted number is $8$
$H=$ getting head on coin
$T=$ be getting tail on coin
$\therefore$ By law of total probability,
$ \begin{aligned} P\left(E _1\right) & =P(H) \cdot P\left(E _1 / H\right)+P(T) \cdot P\left(E _1 / T\right) \\ \text { and } \quad P\left(E _2\right) & =P(H) \cdot P\left(E _2 / H\right)+P(T) \cdot P\left(E _2 / T\right) \\ \text { where, } P(H) & =1 / 2=P(T) \end{aligned} $
$P\left(E _1 / H\right)=$ probability of getting a sum of $7$ on two dice
Here, favourable cases are
$\{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\}$.
$ \therefore \quad P\left(E _1 / H\right)=\frac{6}{36}=\frac{1}{6} $
Also, $\quad P\left(E _1 / T\right)=$ probability of getting 7 numbered card out of $11$ cards
$ =\frac{1}{11} $
$P\left(E _2 / H\right)=$ probability of getting a sum of $8$ on two dice
Here, favourable cases are
$\{(2,6),(6,2),(4,4),(5,3),(3,5)\}$.
$ \therefore \quad P\left(E _2 / H\right)=\frac{5}{36} $
$P\left(E _2 / T\right)=$ probability of getting ’ 8 ’ numbered card out of 11 cards
$ =1 / 11 $
$\therefore \quad P\left(E _1\right)=\frac{1}{2} \times \frac{1}{6}+\frac{1}{2} \times \frac{1}{11}=\frac{1}{12}+\frac{1}{22}=\frac{17}{132}$
and
$ \begin{aligned} P\left(E _2\right) & =(\frac{1}{2}) \times (\frac{5}{36})+\frac{1}{2} \times \frac{1}{11} \\ & =(\frac{1}{2}) (\frac{91}{396})=(\frac{91}{729}) \end{aligned} $
Now, $E _1$ and $E _2$ are mutually exclusive events.
Therefore,
$ P\left(E _1 \text { or } E _2\right)=P\left(E _1\right)+P\left(E _2\right)=\frac{17}{132}+\frac{91}{792}=\frac{193}{792} $
BETA
