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A bag contains $(2 n+1)$ coins. It is known that n of these coins have a head on both sides, whereas the remaining $(\mathrm{n}+1)$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $31 / 42,$ then $\mathrm{n}$ is equal to
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Verified Answer
The correct answer is:
10
Total number of coins $=2 \mathrm{n}+1$
Consider the following events:
$\mathrm{E}_{1}=$ Getting a coin having head on both
sides from the bag.
$\mathrm{E}_{2}=$ Getting a fair coin from the bag
$\mathrm{A}=$ Toss results in a head
Given: $\mathrm{P}(\mathrm{A})=\frac{31}{42}, \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{\mathrm{n}}{2 \mathrm{n}+1}$
and $\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{\mathrm{n}+1}{2 \mathrm{n}+1}$
Then,
$\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)$
$\Rightarrow \frac{31}{42}=\frac{n}{2 n+1} \times 1+\frac{n+1}{2 n+1} \times \frac{1}{2}$
$\frac{31}{42}=\frac{n}{2 n+1}+\frac{n+1}{2(2 n+1)}$
$\Rightarrow \frac{31}{42}=\frac{3 n+1}{2(2 n+1)}$
$\Rightarrow \frac{31}{21}=\frac{3 n+1}{2 n+1}$
$n=10$
Consider the following events:
$\mathrm{E}_{1}=$ Getting a coin having head on both
sides from the bag.
$\mathrm{E}_{2}=$ Getting a fair coin from the bag
$\mathrm{A}=$ Toss results in a head
Given: $\mathrm{P}(\mathrm{A})=\frac{31}{42}, \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{\mathrm{n}}{2 \mathrm{n}+1}$
and $\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{\mathrm{n}+1}{2 \mathrm{n}+1}$
Then,
$\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)$
$\Rightarrow \frac{31}{42}=\frac{n}{2 n+1} \times 1+\frac{n+1}{2 n+1} \times \frac{1}{2}$
$\frac{31}{42}=\frac{n}{2 n+1}+\frac{n+1}{2(2 n+1)}$
$\Rightarrow \frac{31}{42}=\frac{3 n+1}{2(2 n+1)}$
$\Rightarrow \frac{31}{21}=\frac{3 n+1}{2 n+1}$
$n=10$
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