Question

A biased coin has probability of head as 1/3. This coin is tossed 6 times. Find the probability of geeting

Answer 1

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