Probability 3 Question 38
38. 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)
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Solution:
- 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,
$$ P\left(E _1 \text { or } E _2\right)=P\left(E _1\right)+P\left(E _2\right)=\frac{17}{132}+\frac{91}{792}=\frac{193}{792} $$
