Question

six balls numbered 1 to 6 are placed in a bag. some of the balls are grey and some are white. the balls numbered 2 and 5 are grey. the balls numbered 1, 3, 4, and 6 are white. a ball will be selected from the bag at random. the 6 possible outcomes are listed below. note that each outcome has the same probability. complete parts (a) through (c). write the probabilities as fractions. (a) check the outcomes for each event below. then, enter the probability of the event. event a: the selected ball is white event b: the selected ball has an even number on it event a or b: the selected ball is white or has an even number on it event a and b: the selected ball is white and has an even number on it (b) compute the following.

Explanation:

Step1: Determine total number of outcomes

There are 6 balls, so total number of outcomes $n = 6$.

Step2: Calculate probability of Event A

White - numbered balls are 1, 3, 4, 6. So number of favorable outcomes for Event A, $n_A=4$. Probability $P(A)=\frac{n_A}{n}=\frac{4}{6}=\frac{2}{3}$.

Step3: Calculate probability of Event B

Even - numbered balls are 2, 4, 6. So number of favorable outcomes for Event B, $n_B = 3$. Probability $P(B)=\frac{n_B}{n}=\frac{3}{6}=\frac{1}{2}$.

Step4: Calculate probability of Event A or B

Balls that are white or even - numbered are 1, 2, 3, 4, 6. So number of favorable outcomes for Event A or B, $n_{A\cup B}=5$. Probability $P(A\cup B)=\frac{n_{A\cup B}}{n}=\frac{5}{6}$.

Step5: Calculate probability of Event A and B

Balls that are white and even - numbered are 4, 6. So number of favorable outcomes for Event A and B, $n_{A\cap B}=2$. Probability $P(A\cap B)=\frac{n_{A\cap B}}{n}=\frac{2}{6}=\frac{1}{3}$.

Answer:

Event A: Outcomes: 1, 3, 4, 6; Probability: $\frac{2}{3}$
Event B: Outcomes: 2, 4, 6; Probability: $\frac{1}{2}$
Event A or B: Outcomes: 1, 2, 3, 4, 6; Probability: $\frac{5}{6}$
Event A and B: Outcomes: 4, 6; Probability: $\frac{1}{3}$