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.
image of balls: 1 (white), 2 (grey), 3 (white), 4 (white), 5 (grey), 6 (white)
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.
table with outcomes (1,2,3,4,5,6) and probability, and events: event a (white), event b (even), event a or b, event a and b
(b) compute the following.
p(a) + p(b) − p(a and b) = blank

Explanation:

Part (a)

Event A: The selected ball is white

• White balls are numbered 1, 3, 4, 6. So we check these outcomes.
• Number of favorable outcomes for A: 4
• Total number of outcomes: 6
• Probability of A: $\frac{4}{6} = \frac{2}{3}$

Event B: The selected ball has an even number

• Even - numbered balls are 2, 4, 6. So we check these outcomes.
• Number of favorable outcomes for B: 3
• Total number of outcomes: 6
• Probability of B: $\frac{3}{6}=\frac{1}{2}$

Event A or B: The selected ball is white or has an even number

• To find the outcomes for A or B, we consider all outcomes that are in A or in B or in both.
• Outcomes in A: 1, 3, 4, 6; Outcomes in B: 2, 4, 6. The union of these sets is {1, 2, 3, 4, 6} (5 outcomes)
• Number of favorable outcomes for A or B: 5
• Probability of A or B: $\frac{5}{6}$

Event A and B: The selected ball is white and has an even number

• Outcomes that are in both A and B are the white balls with even numbers, which are 4, 6.
• Number of favorable outcomes for A and B: 2
• Probability of A and B: $\frac{2}{6}=\frac{1}{3}$

Part (b)

We know that $P(A)=\frac{2}{3}$, $P(B) = \frac{1}{2}$ and $P(A\text{ and }B)=\frac{1}{3}$

Step 1: Substitute the values into the formula $P(A)+P(B)-P(A\text{ and }B)$

$P(A)+P(B)-P(A\text{ and }B)=\frac{2}{3}+\frac{1}{2}-\frac{1}{3}$

Step 2: Find a common denominator (which is 6) and simplify

$\frac{2\times2}{3\times2}+\frac{1\times3}{2\times3}-\frac{1\times2}{3\times2}=\frac{4}{6}+\frac{3}{6}-\frac{2}{6}$

Step 3: Perform the arithmetic operations

$\frac{4 + 3- 2}{6}=\frac{5}{6}$

Final Answers

Part (a)

• Event A: Probability = $\frac{2}{3}$ (Outcomes 1, 3, 4, 6 are checked)
• Event B: Probability = $\frac{1}{2}$ (Outcomes 2, 4, 6 are checked)
• Event A or B: Probability = $\frac{5}{6}$ (Outcomes 1, 2, 3, 4, 6 are checked)
• Event A and B: Probability = $\frac{1}{3}$ (Outcomes 4, 6 are checked)

Part (b)

$P(A)+P(B)-P(A\text{ and }B)=\frac{5}{6}$

Answer:

Part (a)

Event A: The selected ball is white

• White balls are numbered 1, 3, 4, 6. So we check these outcomes.
• Number of favorable outcomes for A: 4
• Total number of outcomes: 6
• Probability of A: $\frac{4}{6} = \frac{2}{3}$

Event B: The selected ball has an even number

• Even - numbered balls are 2, 4, 6. So we check these outcomes.
• Number of favorable outcomes for B: 3
• Total number of outcomes: 6
• Probability of B: $\frac{3}{6}=\frac{1}{2}$

Event A or B: The selected ball is white or has an even number

• To find the outcomes for A or B, we consider all outcomes that are in A or in B or in both.
• Outcomes in A: 1, 3, 4, 6; Outcomes in B: 2, 4, 6. The union of these sets is {1, 2, 3, 4, 6} (5 outcomes)
• Number of favorable outcomes for A or B: 5
• Probability of A or B: $\frac{5}{6}$

Event A and B: The selected ball is white and has an even number

• Outcomes that are in both A and B are the white balls with even numbers, which are 4, 6.
• Number of favorable outcomes for A and B: 2
• Probability of A and B: $\frac{2}{6}=\frac{1}{3}$

Part (b)

We know that $P(A)=\frac{2}{3}$, $P(B) = \frac{1}{2}$ and $P(A\text{ and }B)=\frac{1}{3}$

Step 1: Substitute the values into the formula $P(A)+P(B)-P(A\text{ and }B)$

$P(A)+P(B)-P(A\text{ and }B)=\frac{2}{3}+\frac{1}{2}-\frac{1}{3}$

Step 2: Find a common denominator (which is 6) and simplify

$\frac{2\times2}{3\times2}+\frac{1\times3}{2\times3}-\frac{1\times2}{3\times2}=\frac{4}{6}+\frac{3}{6}-\frac{2}{6}$

Step 3: Perform the arithmetic operations

$\frac{4 + 3- 2}{6}=\frac{5}{6}$

Final Answers

Part (a)

• Event A: Probability = $\frac{2}{3}$ (Outcomes 1, 3, 4, 6 are checked)
• Event B: Probability = $\frac{1}{2}$ (Outcomes 2, 4, 6 are checked)
• Event A or B: Probability = $\frac{5}{6}$ (Outcomes 1, 2, 3, 4, 6 are checked)
• Event A and B: Probability = $\frac{1}{3}$ (Outcomes 4, 6 are checked)

Part (b)

$P(A)+P(B)-P(A\text{ and }B)=\frac{5}{6}$