NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

A box contains $3$ coins, one coin is fair, one coin is two headed and one coin is weighted, so that the probability of heads appearing is $\frac{1}{3}$ . A coin is selected at random and tossed, then the probability that head appears, is:

Choose the correct answer:

Explanation

Calculation:

Let $E$ be the event that a head appears.

Let $C_1, C_2, \text{ and } C_3$ be the events of selecting a fair coin, a two-headed coin, and a weighted coin, respectively.

Since one coin is selected at random from three:

P(C_1) = P(C_2) = P(C_3) = \frac{1}{3} \text{}

Now, the probability of getting a head for each coin is:

For a fair coin ( $C_1$ ): $P(E|C_1) = \frac{1}{2}$
For a two-headed coin ( $C_2$ ): $P(E|C_2) = 1$
For a weighted coin ( $C_3$ ): $P(E|C_3) = \frac{1}{3}$

Using the Law of Total Probability:

P(E) = P(C_1) \cdot P(E|C_1) + P(C_2) \cdot P(E|C_2) + P(C_3) \cdot P(E|C_3) \text{}

P(E) = \left( \frac{1}{3} \times \frac{1}{2} \right) + \left( \frac{1}{3} \times 1 \right) + \left( \frac{1}{3} \times \frac{1}{3} \right) \text{}

P(E) = \frac{1}{6} + \frac{1}{3} + \frac{1}{9}

To solve, find a common denominator (18):

P(E) = \frac{3 + 6 + 2}{18}

P(E) = \frac{11}{18} \text{}