A box contains a fair coin and a two headed coin. A coin is selected at random and tossed. If head appears, the other coin is tossed. If tail appears, the same coin is tossed. Which of the following statements is/are true?

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The probability that head appear on the second toss is $\frac{5}{8}$

If head appeared on the second toss, then the probability that it also appeared on the first toss is $\frac{4}{5}$

If head appeared on the second toss, then the probability that tail appeared on the first toss is $\frac{2}{5}$

The probability that tail appear on the second toss is $\frac{1}{5}$

answer is A, B.

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Detailed Solution

Question Image
N=Normal coin
D=Double headed coin
E=Head on the $2^{n d}$ toss
$E_{1} =$ Head on the $1^{s t}$ toss
$E_{2} =$ Tail on the $1^{s t}$ toss
$(A) E = (E \cap N) + (E \cap D) = P (N) . P (E / N) + P (D) . P (E / D)$
$\begin{array}{l} = \frac{1}{2} [\frac{1}{2} . \frac{1}{2} + \frac{1}{2} .1] + \frac{1}{2} [1. \frac{1}{2}] \\ T H H H \end{array} = \frac{3}{8} + \frac{2}{8} = \frac{5}{8}$
$(B) P (E_{1} / E) = \frac{P (E_{1} \cap E)}{P (E)} = \frac{\frac{1}{4} + \frac{1}{4}}{\frac{5}{8}} = \frac{4}{5}$
$(C) P (E_{2} / E) = \frac{P (E_{2} \cap E)}{P (E)} = \frac{\frac{1}{2} (\frac{1}{2} \times \frac{1}{2})}{\frac{5}{8}} = \frac{1}{5}$
$\begin{array}{l} (D) \frac{1}{2} (\frac{1}{2} \times \frac{1}{2}) = \frac{1}{8} \\ T T \end{array}$