Skip to content
doubtly

1 answer

Best answer

Answer: \(P(A) = \dfrac{1}{4}\), \(P(B) = \dfrac{1}{4}\)

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}, n(S) = 12

Event A: head and a prime number (2, 3, 5). A = {H2, H3, H5}, n(A) = 3

\(P(A) = \dfrac{3}{12} = \dfrac{1}{4}\)

Event B: tail and an odd number (1, 3, 5). B = {T1, T3, T5}, n(B) = 3

\(P(B) = \dfrac{3}{12} = \dfrac{1}{4}\)

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report
Was this answer helpful?

Your answer

Log in or create a free account to answer this question.