A bag contains n+1 coins. It is known that one of these coins shows...

A bag contains

n+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

\frac{7}{12},

then the value of

n

is.

Options

A3
B4
C5
DNone of these

Correct answer

C. 5

Step-by-step solution

Let

E_{1}

denote the event "a coin with head on both sides is selected" and

\mathrm{E}_{2}

denotes the event " a fair coin is selected". Let A

be the event " he toss, results in heads".

\begin{array}{l} \therefore\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{n+1}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{n}{n+1} \text{ and }\\ \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{1}}\right)=1, \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)=\frac{1}{2}\\ \therefore \mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{1}}\right)+\mathrm{P} \left(\mathrm{E}_{2}\right) \mathrm{P}\ left(\frac{\mathrm{A}}{\mathrm{E}_{2}}) \\ \Rightarrow \frac{7}{12}=\frac{1}{n+1} \t ttimes₁⁺\frac{n}{n+1} \times \frac{1}{2} \\ \Rightarrow 14 n+14=24+12 n \Rightarrow n=5 \end{array}

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