A biased coin is tossed 10 times. The head is 2 times more likely to appear than the tail. The probability that 2 n d tail and 4 t h tail occur at 4 t h and 10 t h tosses respectively is

Step-by-step solution

P H = 2 3 , P T = 1 3 2 n d tail and 4 t h tail occur at 4 t h and 10 t h tosses respectively implies that there is one tail in first three tosses, 2 n d tail at 4 t h toss, one tail at next five tosses and 4 t h tail at 10 t h toss. So, the required probability = C 3 1 1 3 2 3 2 × 1 3 × C 5 1 1 3 1 2 3 4 × 1 3 = 320 3 9