Probability 3 Question 5
5. An unbiased coin is tossed. If the outcome is a head, then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail, then a card from a well-shuffled pack of nine cards numbered $1,2,3, \ldots, 9$ is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is
(a) $\frac{15}{72}$
(b) $\frac{13}{36}$
(c) $\frac{19}{72}$
(d) $\frac{19}{36}$
(2019 Main, 10 Jan I)
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Solution:
- Clearly, $P(H)=$ Probability of getting head $=\frac{1}{2}$
$$ \text { and } \quad P(T)=\text { Probability of getting tail }=\frac{1}{2} $$
Now, let $E _1$ be the event of getting a sum 7 or 8 , when a pair of dice is rolled.
Then, $E _1={(6,1),(5,2),(4,3),(3,4),(2,5)$,
$(1,6),(6,2),(5,3),(4,4),(3,5),(2,6)}$
$\Rightarrow P\left(E _1\right)=$ Probability of getting 7 or 8 when a pair of dice is thrown $=\frac{11}{36}$
Also, let $P\left(E _2\right)=$ Probability of getting 7 or 8 when a card is picked from cards numbered $1,2, \ldots ., 9=\frac{2}{9}$
$\therefore$ Probability that the noted number is 7 or 8
$$ \begin{aligned} & =P\left(\left(H \cap E _1\right) \text { or }\left(T \cap E _2\right)\right) \\ & =P\left(H \cap E _1\right)+P\left(T \cap E _2\right) \end{aligned} $$
$\left[\because\left(H \cap E _1\right)\right.$ and $\left(T \cap E _2\right)$ are mutually exclusive] $=P(H) \cdot P\left(E _1\right)+P(T) \cdot P\left(E _2\right)$
$\left[\because{H, E _1 }\right.$ and ${T, E _2 }$ both are sets of $=\frac{1}{2} \times \frac{11}{36}+\frac{1}{2} \times \frac{2}{9}=\frac{19}{72}$ independent events]
