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 \(\dfrac{1}{3}\) . A coin is selected at random and tossed, then the probability that head appears, is
Case A: Selecting a Coin \[ P(A) = \frac{1}{3} \] Case B: Head Appears \[ P(B) = \frac{1}{2} \text{ (for fair coin) } + 1 \text{ (for two-headed coin) } + \frac{1}{3} \text{ (for biased coin)} \] \[ \Rightarrow P(B) = \frac{11}{6} \] The Probability That a Head Appears \[ P(X) = P(A) \times P(B) \] \[ \Rightarrow \mathbf{P(X)} = \frac{1}{3} \times \frac{11}{6} = \frac{11}{18} \]
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