A fair die is rolled nine times. What is the probability that

anodd number (1,3 or 5) will occur less than 3 times?

The answer is 46/512 = 0.08984375

Step-by-step explanation:

In this question, success means to get an odd number on rolling a die.

The probability of success p = 3/6 = 1/2

So p = 1/2

Now lets find probability of failure q

q = 1 - p = 1 - 1/2 = 1/2

So q = 1/2

Let X = x denotes the number of success in n trials.

So X is a Binomial Random Variable

Which has the following parameters:

n = 9

as the fair die is rolled nine times and

p = 1/2

Now the Probability of x success out of n trials P (X = x) is:

P (X = x) = p (x) = nCx pˣ, qⁿ⁻ˣ, x = 0,1,2, ...,9

P (X = x) = p (x) = nCx (1/2) ⁹ = ₉Cₓ (1/2) ⁹ = ₉Cₓ / 512

Since the required probability is P (X < 3) So

P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)

= 1 / 512 {₉C₀ + ₉C₁ + ₉C₂} nCr = n! / r! * (n - r) !

= 1 / 512{ (9! / 0! * (9 - 0) !) + (9! / 1! * (9 - 1) !) + (9! / 2! * (9 - 2) !) }

= 1 / 512 { 1 + 9 + 36 }

= 46 / 512 = 0.08984375

So the required probability is 46/512