Question

1) Suppose that three balls are sampled without replacement from an urn containing 5 red balls and 5 green balls. Find the probability that the sample contains at least one ball of each color 2.) Suppose that a pair of fair dice is rolled and that all 36 outcomes are equally likely. Calculate the probability that the second die lands on a number greater than the first.

Answer 1

1.

The probability of choosing only one color ball from the urn is

$\frac{\operatorname{nCr}\left(5,3\right)}{\operatorname{nCr}\left(10,3\right)}$

We have two such cases possible because we have two colors

So the probability of choosing any color ball from the urn is

$2* \frac{\operatorname{nCr}\left(5,3\right)}{\operatorname{nCr}\left(10,3\right)}$

And the probability of not choosing only one colored ball is

$1-2\cdot\frac{\operatorname{nCr}\left(5,3\right)}{\operatorname{nCr}\left(10,3\right)}$

= 0.8333

2.

66。 666 1-23456 5 1-23456 4 4 5-6 2 3 123456 2 1-23456 1 123456 123456

Second die lands on a number greater than the first one are given in the table.

The count of those are 15

So the probability is 15/36= 0.416666