Find the probability of rolling at least 3 even numbers in 7 rolls of six sided die

Question

I need help setting this problem up.

Answer 1

Hello, thank you for taking the time to post your question!

You can think of this scenario as being a binomial probability since we have a fixed number of trails (n = 7 rolls), each trial is independent, there are only two outcomes, and the probability of success (p = 0.50) is constant.

So the parameter that you’re testing then is that you want x >= 3 since it says it wants you to find the probability of rolling at least 3 even numbers.

When I plug in those underlying parameters into a binomial calculator I get 0.7734 for the end value on this one! So the probability of rolling at least 3 even numbers in this situation is 0.7734.

I hope that helps get you moving in the right direction! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

Hi Kaylee,

 

This is a binomial probability problem. You have two outcomes (that you're interested in) for each roll of the die: if it's even or odd. The probability of rolling an even number never changes, and you have a fixed number of rolls (7).

 

The general formula for finding the probability of x successes out of n events is:

P(X) = _nC_x p^x(1-p)^n-x

 

_nC_x = n! / x!(n-x)!

 

In your problem,

n = 7,

p = 3/6 = 0.50 (chance of rolling an even number)

(1-p) = 1 - 0.50 = 0.50

 

You want to find the probability of rolling at least 3 even numbers. The possible number of even numbers you roll can be anywhere between 0 and 7, and you want those that are 3 or more.

0 1 2 3 4 5 6 7

 

 

To find this, you can either calculate

P(X≥3) = P(3)+P(4)+P(5)+P(6)+P(7)

or

P(X≥3) = 1 - P(X<3) = 1 - [P(0)+P(1)+P(2)]

 

You calculate each of these using the formula above. For example,

P(3) = ₇C₃ (0.5)³(0.5)⁴

P(3) = 35 * (0.5)⁷ = 0.2734

 

I hope this helps. If you have further questions, please comment and ask.