A die is thrown. If it shows 1 or 2, a coin is tossed once; otherwise, the coin is tossed twice. Let H be the event that exactly one head is obtained. Consider the following statements:
(A) \(P(H) = \displaystyle\frac{1}{2}\).
(B) The probability that the die showed 1 or 2, given H, is \(\displaystyle\frac{1}{3}\).
(C) The probability that no head is obtained is \(\displaystyle\frac{1}{4}\).
(D) The probability that the coin was tossed twice, given that no head is obtained, is \(\displaystyle\frac{1}{2}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
Let \(E_1\) be the event that the die shows 1 or 2 (one toss) and \(E_2\) the event that it shows 3, 4, 5 or 6 (two tosses).
\( \displaystyle P(E_1) = \frac{1}{3},\quad P(E_2) = \frac{2}{3} \)
With one toss, exactly one head has probability \(\frac{1}{2}\) and no head has probability \(\frac{1}{2}\). With two tosses, exactly one head (HT or TH) has probability \(\frac{2}{4} = \frac{1}{2}\) and no head (TT) has probability \(\frac{1}{4}\).
By the theorem of total probability:
\( \displaystyle P(H) = \frac{1}{3}\times\frac{1}{2} + \frac{2}{3}\times\frac{1}{2} = \frac{1}{2} \)
\( \displaystyle P(\text{no head}) = \frac{1}{3}\times\frac{1}{2} + \frac{2}{3}\times\frac{1}{4} = \frac{1}{6}+\frac{1}{6} = \frac{1}{3} \)
By Bayes’ theorem:
\( \displaystyle P(E_1|H) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3} \)
\( \displaystyle P(E_2|\text{no head}) = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2} \)
So P(H) = \(\frac{1}{2}\), P(one toss | H) = \(\frac{1}{3}\), P(no head) = \(\frac{1}{3}\) and P(two tosses | no head) = \(\frac{1}{2}\).
Checking the statements
(A) \(P(H) = \frac{1}{2}\). — Correct. \(\frac{1}{3}\times\frac{1}{2}+\frac{2}{3}\times\frac{1}{2} = \frac{1}{2}\).
(B) The probability that the die showed 1 or 2, given H, is \(\frac{1}{3}\). — Correct. \(\frac{1}{6}\div\frac{1}{2} = \frac{1}{3}\).
(C) The probability that no head is obtained is \(\frac{1}{4}\). — Incorrect. It is \(\frac{1}{6}+\frac{1}{6} = \frac{1}{3}\); \(\frac{1}{4}\) is the value for two tosses only.
(D) The probability that the coin was tossed twice, given that no head is obtained, is \(\frac{1}{2}\). — Correct. \(\frac{1}{6}\div\frac{1}{3} = \frac{1}{2}\).