A fair die is rolled nine times. What is the probability that an odd number (1,3 or 5) will occur less than 3?

Question

A fair die is rolled nine times. What is the probability that an odd number (1,3 or 5) will occur less than 3 times?

Answer 1

Let X be the number of times an odd number is rolled.  X has the binomial distribution with n = 9 trials and success probability p = 0.5



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n = 9 , p = 0.5 )

the mean of the binomial distribution is n * p = 4.5

the variance of the binomial distribution is n * p * (1 - p) = 2.25

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.5

The Probability Mass Function, PMF,

f(X) = P(X = x) is:

P( X =  0 ) =  0.001953125

P( X =  1 ) =  0.01757813

P( X =  2 ) =  0.0703125

P( X =  3 ) =  0.1640625

P( X =  4 ) =  0.2460938

P( X =  5 ) =  0.2460938

P( X =  6 ) =  0.1640625

P( X =  7 ) =  0.0703125

P( X =  8 ) =  0.01757813

P( X =  9 ) =  0.001953125

The Cumulative Distribution Function, CDF,

F(X) = P(X ≤ x) is:

x

∑ P(X = t) =

t = 0

P( X ≤ 0 ) =  0.001953125

P( X ≤ 1 ) =  0.01953125

P( X ≤ 2 ) =  0.08984375 = P(X < 3) <<< ANSWER

P( X ≤ 3 ) =  0.2539063

P( X ≤ 4 ) =  0.5

P( X ≤ 5 ) =  0.7460938

P( X ≤ 6 ) =  0.9101563

P( X ≤ 7 ) =  0.9804688

P( X ≤ 8 ) =  0.9980469

P( X ≤ 9 ) =  1