Question

A fair coin is tossed ten times. What is the probability of obtaining at most nine heads?

Answer 1

Let X denote the numbe of heads when coin is tossed ten times.

Probability of obtaining at most nine heads = P(X 9 )

Using Sum of probabilites rule,

P(X 9 ) = 1 - P(X=10)

Here, X will follow Binomial distribution as :

where,

n = Number of trials

p = Probability of success

x = Number of success

In our case,

n = 10

p = Probability oh heads = 1/2 = 0.5

Required probability = 1 - P(X=10)

Note :

n! = n*(n-1)*(n-2)*......................*1

Solving this we get ,

P(X=10) = 0.000977

Probability of obtaining at most nine heads = 1 - P(X=10) = 1 - 0.000977 = 0.999023