Question

A fair die will be rolled 11 times. What is the probability that an odd number is rolled exactly 7 times?

Answer 1

0.161

Explanation:

Given:

Number of trials,
`n=11`

Consider the event of rolling odd number as success. There are 3 odd and 3 even numbers in a fair die.

So, probability of success,
`p=0.5`

Probability of failure,
`q=1-p=1-0.5=0.5`

Number of successes,
`x=7`

From Bernoulli's Theorem, the probability of
`x` successes in
`n` trials is given as,

`P(X=x)=_(x)^(n)\textrm{C}p^(x)q^(n-x)`

Here,
`x=7,n=11,p=0.5,q=0.5`

So,
`P(X=7)=_(7)^(11)\textrm{C}(0.5)^(7)q^(11-7)\\ P(X=7)=0.161`

Therefore, the probability of getting an odd number exactly 7 times is 0.161.