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Q.

A box contain three coins; one coin is fair, one coin is two-headed, and one coin is weighted so that the probability of heads, appearing is 1/3. A coin is selected at random and tossed. Then the probability that heads appears is

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a

$\frac{11}{18}$

b

$\frac{12}{18}$

c

$\frac{13}{18}$

d

$\frac{14}{18}$

answer is A.

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Detailed Solution

$E_{1} =$ fair coin selected

$E_{2} =$ two headed coin selected

$E_{3} =$ selected weighted coin

$P (E_{1}) = \frac{1}{3} P (E_{2}) = \frac{1}{3} P (E_{3}) = \frac{1}{3}$

A=getting head

$\begin{array}{rcl} P (\frac{A}{E_{1}}) & = & \frac{1}{2} P (\frac{A}{E_{2}}) = 1 P (\frac{A}{E_{3}}) = \frac{1}{3} \\ P (A) & = & \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times 1 + \frac{1}{3} \times \frac{1}{3} = \frac{3 + 6 + 2}{18} \\ = & \frac{11}{18} \end{array}$

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