A biased coin that has a probability of getting heads as p ( 0 < p < 1 ) is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is 2 5 , then 9 p is equal to

Step-by-step solution

p : probability of getting head q : probability of getting tail Such that, p + q = 1 According to question we can get head in 2 n d , &#160; 4 t h , &#160; 6 t h &hellip; tail. Hence required probability = q p 1 + q 2 + q 4 + &#8230; = 2 5 &#8658; q p 1 - q 2 = 2 5 &#8658; 5 p q = 2 &#8722; 2 q 2 &#8658; 5 p ( 1 &#8722; p ) = 2 &#8722; 2 ( 1 &#8722; p ) 2 &#8658; 5 p &#8722; 5 p 2 = 2 &#8722; 2 ( 1 + p 2 &#8722; 2 p ) &#8658; 5 p &#8722; 5 p 2 = 2 &#8722; 2 &#8722; 2 p 2 + 4 p &#8658; 3 p 2 = p &#8658; p ( 3 p &#87