List-I | List-II (Adverbs) |
(A) P(exactly 2 heads) | (I) \(\frac{1}{4}\) |
(B) P(at least 1 head) | (II) \(1\) |
(C) P(at most 2 heads) | (III) \(\frac{3}{4}\) |
(D) P(exactly 1 head) | (IV) \(\frac{1}{2}\) |
(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(A): $P(exactly 2 heads)$ occurs when both tosses are heads (HH). The probability is:
$P(exactly 2 heads) = \frac{1}{4}$.
Thus, (A)-(I).
(B): $P(at least 1 head)$ occurs in the cases HT, TH, and HH. The probability is:
$P(at least 1 head) = 1 - P(no heads) = 1 - \frac{1}{4} = \frac{3}{4}$.
Thus, (B)-(III).
(C): $P(at most 2 heads)$ includes all possible outcomes (HH, HT, TH, TT). The probability is:
$P(at most 2 heads) = 1$.
Thus, (C)-(II).
(D): $P(exactly 1 head)$ occurs in the cases HT and TH. The probability is:
$P(exactly 1 head) = \frac{2}{4} = \frac{1}{2}$.
Thus, (D)-(IV).
LIST-I(EVENT) | LIST-II(PROBABILITY) |
(A) The sum of the number is greater than 11 | (i) 0 |
(B) The sum of the number is 4 or less | (ii) 1/15 |
(C) The sum of the number is 4 | (iii) 2/15 |
(D) The sum of the number is 4 | (iv) 3/15 |
Choose the correct answer from the option given below