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Question

Mathematics Question on Random Experiments

A bag contains n+1n + 1 coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is 712\frac{7}{12}, then the value of nn is.

A

3

B

4

C

5

D

None of these

Answer

5

Explanation

Solution

Let E1E_1 denote the event "a coin with head on both sides is selected" and E2E_2 denotes the event " a fair coin is selected".
Let AA be the event " he toss, results in heads".
P(E1)=1n+1,P(E2)=nn+1\therefore P\left(E_{1}\right) = \frac{1}{n+1} , P\left(E_{2}\right) = \frac{n}{n+1} and
P(AE1)=1,P(AE2)=12P\left(\frac{A}{E_{1}} \right) = 1, P\left(\frac{A}{E_{2}}\right) = \frac{1}{2}
P(A)=P(E1)P(AE1)+P(E2)P(AE2)\therefore P\left(A\right) =P\left(E_{1}\right)P\left(\frac{A}{E_{1}}\right) + P\left(E_{2}\right)P\left(\frac{A}{E_{2}}\right)
712=1n+1×1+nn+1×12\Rightarrow \frac{7}{12} = \frac{1}{n +1} \times1 + \frac{n}{n+1} \times \frac{1}{2}
14n+14=24+12n\Rightarrow 14n +14 = 24 +12n
n=5\Rightarrow n = 5