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Answer» It seems your question is incomplete . complete question may be -------> A BIASED coin has probability of head as 1/3. This coin is tossed 6 times. Find the probability of getting at least two heads ?
SOLUTION :- probability of getting at least two heads = 1 - probability of getting at most one head
Given, probability of head , P(head) = 1/3 probability of TAIL , P(Tail) = 1 - P(head) = 2/3 Now , probability of getting at most head = probability of getting one head back+ probability of getting ZERO head e.g., PX ≤ 1) = P(x = 1) + P(x = 0) = ⁶C₆₋₁(2/3)⁶⁻¹(1/3)¹ + ⁶C₆(2/3)⁶(1/3)⁰ = ⁶C₅ (2/3)⁵ (1/3) + 1 × (2/3)⁶ × 1 = 2 × 32/243 + 64/729 = 64/243 + 64/729 = (64 × 3 + 64)/729 = 256/729
∴ probability of getting two heads = 1 - P(X≤1) = 1 - 256/729 = (729 - 256)/729 = 473/729
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