Question:

A coin is tossed twice. Then Match List-I with List-II:
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) \(\frac{1}{2}\)

Choose the correct answer from the options given below :

Updated On: Nov 28, 2024
  • (A)-(I), (B)-(III), (C)-(II), (D)-(IV)

  • (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  • (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  • (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

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The Correct Option is A

Solution and Explanation

(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).

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