In tossing a fair coin, what is
the probability that the second
"heads” occurs on the ninth toss?

1/64

Step-by-step explanation:

We want a second H in the 9th toss. So, we must get only 1 H in the first eight tosses.

So, in the first 8 tosses, we have 1 H and 7 T. Number of ways of getting 1 H and 7 T is 8!/7! = 8

Total possible outcomes = 2^9 = 512

Required probability = 8/512 = 1/64

[ If there is any problem with the solution, please inform me ]

probability for j is 1/4

while

a is 1/6

table a is wierd

step-by-step explanation:

6*f