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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Probability

BITSAT 2015 · Mathematics · Probability

A bag contains (2 n+1) coins. It is known that n of these coins...

A bag contains (2n+1)(2 n+1) coins. It is known that n of these coins have a head on both sides, whereas the remaining (n+1)(\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 31/42,31 / 42, then n\mathrm{n} is equal to

Options

  1. A10
  2. B11
  3. C12
  4. D13

Correct answer

A. 10

Step-by-step solution

Total number of coins =2n+1=2 \mathrm{n}+1 Consider the following events: E1=\mathrm{E}_{1}= Getting a coin having head on both sides from the bag. E2=\mathrm{E}_{2}= Getting a fair coin from the bag A=\mathrm{A}= Toss results in a head Given: P(A)=3142,P(E1)=n2n+1\mathrm{P}(\mathrm{A})=\frac{31}{42}, \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{\mathrm{n}}{2 \mathrm{n}+1} and P(E2)=n+12n+1\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{\mathrm{n}+1}{2 \mathrm{n}+1} Then, P(A)=P(E1)P(A/E1)+P(E2)P(A/E2)⇒3142=n2n+1×1+n+12n+1×123142=n2n+1+n+12(2n+1)⇒3142=3n+12(2n+1)⇒3121=3n+12n+1n=10\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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