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  • A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is 31|42, deter

A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is 31/        42, determine the value of n.

Answers (1)

Given-

n coins have head on both the sides and (n + 1) coins are fair coins

Therefore, Total coins = 2n + 1
Let E1, E2 be the following events:

E1 = Event that an unfair coin is selected
E2 = Event that a fair coin is selected

\therefore P\left(E_{1}\right)=\frac{n}{2 n+1} \text { and } P\left(E_{2}\right)=\frac{n+1}{2 n+1}

The Law of Total Probability:


In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then
P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)
Let “E” be the event that the toss result is a head
P(E|E1) is the probability of getting a head when unfair coin is tossed
P(E|E2) is the probability of getting a head when fair coin is tossed
Therefore,

\begin{aligned} &P\left(E \mid E_{1}\right)=1 \text { and } P\left(E \mid E_{2}\right)=\frac{1}{2}\\ &\text { Erom the law of total probability, }\\ &\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{E} \mid \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} \text { (Given) } \end{aligned}

\begin{aligned} &\Rightarrow \frac{31}{42}=\frac{2 n+n+1}{2(2 n+1)}\\ &\Rightarrow 31 \times 2(2 n+1)=42 \times(3 n+1)\\ &\Rightarrow 124 n+62=126 n+42\\ &\Rightarrow 2 n=20\\ &\Rightarrow \mathrm{n}=10\\ &\text { Hence, the value of } \mathrm{n} \text { is } 10 \text { . } \end{aligned}

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