Question

An urn contains 11 white balls and 9 green balls. A sample of seven is selected at random. What is the probability that the sample contains at least one green ball

Answer 1

Explanation:

Note that the question wants the chance of at LEAST one green ball. that means containing 1,2,3,4... etc is all fair game. Now calculating all of those odds by hand is a pain. But we dont have to. There's a clever little trick we can use to cut time: The sum of the odds of all the possible events (no greens, 1 green, etc) must be 100%. Knowing that, we can just calculate the odds of getting no green balls and subtract that from 100%.

The odds of no green balls are:

`(11)/(20) \cdot(10)/(19) \cdot(9)/(18) \cdot(8)/(17) \cdot...\\`
`\cdot (4)/(13)`

Wich can be rewritten as:

`(11!12!)/(20!3!) \approx 0,0013` or 0,13%

Subtracting from 100%:

100-0,13=99,87% of at least one green