Calculating Chances of Success with Binomial Distribution: A Homework Example

[Schmidtter]

Homework Statement

I am trying to figure out if I have a 20% chance to get what I want (let it = x) and I have 6 chances to do so (n=6), I am curious how I set this question up to find out my chances of getting 'x' once out of the 6 times I try.

Homework Equations

Binomial Distribution.

x! / y!(x - y)!

The Attempt at a Solution

x = 0.2
y = 0.8

= 0.8! / 0.2! (0.8 - 0.2)!
= 40320 / 1440
= 28

Any help would be appreciated.

 

[Dick]

I don't think you quite get the binomial distribution. What is '.8!' suppose to mean? You might want to review it. On the other hand, the easy way to solve this problem is to figure out your odds of losing 6 straight times, call it P. Then your odds of winning are 1-P.

 

[Architeuthis Dux]

I would suggest using the binomial distribution formula to calculate the probability of getting 'x' once out of 6 tries. The formula is P(x) = nCx * px * (1-p)n-x, where n is the number of trials, p is the probability of success, and x is the number of successes. In this case, n = 6, p = 0.2, and x = 1. Plugging these values into the formula, we get P(x) = 6C1 * (0.2) * (1-0.2)6-1 = 6 * 0.2 * 0.8^5 = 0.3932. This means that there is a 39.32% chance of getting 'x' once out of 6 tries with a 20% chance of success each time.