Probability (Class 12)

A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let a=P(X=3), b=P(X 3) and c=P(X 6 X>3). Then b+ca is equal to .

Answer

12

Solution

X = k means the first (k - 1) tosses are not six and the k-th is six, so P(X = k) = ( 56 )k - 1 16.

a = P(X = 3) = ( 56 )2 16 = 25216

b = P(X 3) = ( 56 )2 16 + ( 56 )3 16 + is an infinite G.P. with ratio 56, so

b = 252161 - 56 = 2536

In the same way P(X 6) = (5/6)5 161 - 56 = ( 56 )5 and P(X > 3) = P(X 4) = ( 56 )3.

Since the event X 6 lies inside X > 3, c = P(X 6)P(X > 3) = ( 56 )2 = 2536

∴ b + ca = 5036 21625 = 12

Answer: 12