A pair of coins thrown then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | Probability of occurrence of head on both coins is | I. | \( \displaystyle\frac{3}{4} \) |
| B. | Probability of occurrence of at least one tail is | II. | \( \displaystyle\frac{1}{2} \) |
| C. | Probability of occurrence of one tail only is | III. | 1 |
| D. | Probability of a tail appears on one coin given that only one coin shows head. | IV. | \( \displaystyle\frac{1}{4} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-I, C-II, D-III
The sample space is S = {HH, HT, TH, TT}, with 4 equally likely outcomes.
(A) Probability of occurrence of head on both coins
This matches (IV) \( \frac{1}{4} \). Only HH is favourable.
(B) Probability of occurrence of at least one tail
This matches (I) \( \frac{3}{4} \). HT, TH and TT are favourable; equally, 1 − P(HH) = 1 − \( \frac{1}{4} \).
(C) Probability of occurrence of one tail only
This matches (II) \( \frac{1}{2} \). HT and TH are favourable, 2 out of 4.
(D) Probability of a tail on one coin, given that only one coin shows head
This matches (III) 1.
Let F = exactly one head = {HT, TH} and E = a tail on one coin. Every outcome of F has a tail, so E ∩ F = F:
\( \displaystyle P(E\,|\,F)=\frac{P(E\cap F)}{P(F)} \)
\( \displaystyle =\frac{2/4}{2/4}=1 \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).