A biased coin has a probability p of showing Head and q of showing Tail. When this coin is tossed 4 times, the probability of getting exactly 3 Heads is twice the probability of getting exactly 2 Heads. The probability of getting exactly 4 Heads when the coin is tossed 4 times is

Step-by-step solution

Let p be the probability of getting a Head and q be the probability of getting a Tail in a single toss. We know that p + q = 1 . Given that in 4 tosses, the probability of getting exactly 3 Heads is twice the probability of getting exactly 2 Heads: P(X = 3) = 2 P(X = 2) Using the binomial distribution formula: ⁴C₃ p³ q¹ = 2 ( ⁴C₂ p² q² ) Substituting the values of the binomial coefficients: 4 p³ q = 2 6 p² q² 4 p³ q = 12 p² q² Since p > 0 and q > 0 , we can divide both sides by 4 p² q : p = 3q Substitute q = 1 - p