Question

An urn contains five red, three orange, and two blue balls. Two balls are randomly selected. What is the sample space of this experiment? Let \(X\) represent the number of orange balls selected. What are the possible values of \(X\) ? Calculate \(P[X=0\\}\).

Short answer

The sample space of the experiment is the different combinations of colors of balls that can be selected: (Red, Red), (Red, Orange), (Red, Blue), (Orange, Red), (Orange, Orange), (Orange, Blue), (Blue, Red), (Blue, Orange), and (Blue, Blue). The possible values of the random variable \(X\), representing the number of orange balls selected, are 0, 1, or 2. The probability of \(P[X=0]\), which corresponds to selecting no orange balls, is \(\frac{11}{45}\).

Step-by-step solution

Define the sample space

The sample space is the set of all possible outcomes of an experiment, in this case, the different combinations of colors of balls that can be selected. Since we are choosing 2 balls without replacement, the possible outcomes in the sample space are (Red, Red), (Red, Orange), (Red, Blue), (Orange, Red), (Orange, Orange), (Orange, Blue), (Blue, Red), (Blue, Orange), and (Blue, Blue).

Determine the possible values of the random variable \(X\)

The random variable \(X\) is defined as the number of orange balls selected in the experiment. Since we are selecting two balls, the possible values for \(X\) are 0, 1, or 2, i.e., no orange balls, one orange ball, or two orange balls.

Calculate the probability of \(P[X=0]\)

To calculate the probability of \(P[X=0]\), we need to find the fraction of outcomes in the sample space with no orange balls selected. When no orange balls are selected, we have three possible outcomes: (Red, Red), (Red, Blue), and (Blue, Red), (Blue, Blue). Calculate the total number of outcomes by using combinations, since the order of selection does not matter: Total combinations = C(5+3+2, 2) = C(10, 2) = \(\frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = 45\) Now, let's find the number of combinations with no orange balls (i.e., only red and blue balls are selected): Red & Blue combinations = C(5, 2) + C(2, 2) = 10 + 1 = 11 Finally, the probability of selecting no orange balls is given by: \(P[X=0] = \(\frac{\text{Number of combinations with no orange balls}}{\text{Total number of combinations}}\) = \frac{11}{45}\)