Answer to Question #86658 in Statistics and Probability for bob

Question

A box contains five balls numbered 1, 2, 2, 3, and 4. We will randomly select two balls from the box without replacement.

The outcome of interest is the number on each of the two balls we select. How many outcomes are contained in the sample space?
Order matters, so the outcome "12" is a different outcome than "21"

Let X be the sum of the numbers on the two selected balls. Find the probability that X = 3.

Let X be the sum of the numbers on the two selected balls. What is the probability that the first selection is not 1 if we know that X = 4? Report your answer as a decimal.

Answer 1

$$S=\lbrace 12, 13, 14, 21, 22, 23, 24, 31, 32, 34, 41, 42, 43\rbrace$$

The sum of the numbers on the two selected balls is 3

$$\lbrace 12, 21\rbrace$$

Then

$$P(X=3)=2/13=0.153846$$

We know that X = 4

$$\lbrace 13, 22, 31\rbrace$$

The probability that the first selection is not 1 if we know that X = 4 is

$$P(first =\not 1 | X=4)=2/3=0.666667$$