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Question
Two coins are tossed simultaneously. Find the probability of getting at least one head.
Sum
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Solution 1
S = {HH, HT, TH, TT}
n(S) = 4
A = {HH, HT, TH}
n(A) = 3
`P(A) = (n(A))/(n(S))`
= `3/4`
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Solution 2
Given: Two fair coins are tossed simultaneously. Sample space S = {HH, HT, TH, TT} 4 equally likely outcomes.
Step-wise calculation:
1. “At least one head” outcomes = {HH, HT, TH} → count = 3.
2. Total outcomes = 4.
3. Probability = `3/4`.
Alternatively P(at least one head) = 1 – P(no head)
= `1 - 1/4`
= `3/4`
P(at least one head) = `3/4`.
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Is there an error in this question or solution?
Chapter 22: Probability - Exercise 22A [Page 505]
