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Question: Answered & Verified by Expert
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
MathematicsProbabilityBITSATBITSAT 2015
Options:
  • A 10
  • B 11
  • C 12
  • D 13
Solution:
2946 Upvotes 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,

$\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)$

$\Rightarrow \frac{31}{42}=\frac{n}{2 n+1} \times 1+\frac{n+1}{2 n+1} \times \frac{1}{2}$

$\frac{31}{42}=\frac{n}{2 n+1}+\frac{n+1}{2(2 n+1)}$

$\Rightarrow \frac{31}{42}=\frac{3 n+1}{2(2 n+1)}$

$\Rightarrow \frac{31}{21}=\frac{3 n+1}{2 n+1}$

$n=10$

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