Question

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, 7) and probability column; events: event a (number 2–5), event b (grey), event a and b, event a or b
(b) compute the following.
$p(a) + p(b) - p(a \text{ and } b) = \square$
(c) select the answer that makes the equation true.
$p(a) + p(b) - p(a \text{ and } b) =$ select dropdown

Explanation:

To solve this problem, we analyze each part step by step. First, we determine the number of favorable outcomes for each event and then calculate the probabilities. Finally, we use the principle of inclusion - exclusion for probability.

Part (a)

We assume that each outcome (numbered 1 - 7) is equally likely, so the total number of possible outcomes \(n = 7\).

Event A: The selected ball has a number from 2 to 5

The numbers from 2 to 5 are 2, 3, 4, 5. So the number of favorable outcomes \(n_A=4\).
The probability of event A, \(P(A)=\frac{n_A}{n}=\frac{4}{7}\)

Event B: The selected ball is grey

From the table, the grey - colored outcomes are 1, 4, 5, 7. So the number of favorable outcomes \(n_B = 4\).
The probability of event B, \(P(B)=\frac{n_B}{n}=\frac{4}{7}\)

Event A and B: The selected ball has a number from 2 to 5 and is grey

The numbers from 2 to 5 that are grey are 4, 5. So the number of favorable outcomes \(n_{A\cap B}=2\).
The probability of \(A\cap B\), \(P(A\cap B)=\frac{n_{A\cap B}}{n}=\frac{2}{7}\)

Event A or B: The selected ball has a number from 2 to 5 or is grey

Using the principle of inclusion - exclusion, \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). But we can also count the number of favorable outcomes directly. The outcomes in \(A\cup B\) are 1, 2, 3, 4, 5, 7. So \(n_{A\cup B}=6\).
The probability of \(A\cup B\), \(P(A\cup B)=\frac{n_{A\cup B}}{n}=\frac{6}{7}\)

Part (b)

We know that \(P(A)=\frac{4}{7}\), \(P(B)=\frac{4}{7}\) and \(P(A\cap B)=\frac{2}{7}\)
\[

$$\begin{align*} P(A)+P(B)-P(A\cap B)&=\frac{4}{7}+\frac{4}{7}-\frac{2}{7}\\ &=\frac{4 + 4-2}{7}\\ &=\frac{6}{7} \end{align*}$$

\]

Part (c)

From the principle of inclusion - exclusion in probability, we know that \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). From part (a), we found that \(P(A\cup B)=\frac{6}{7}\) and from part (b) we calculated \(P(A)+P(B)-P(A\cap B)=\frac{6}{7}\). So \(P(A)+P(B)-P(A\cap B)=P(A\cup B)\)

Final Answers

Part (a)

• Event A: \(\frac{4}{7}\)
• Event B: \(\frac{4}{7}\)
• Event A and B: \(\frac{2}{7}\)
• Event A or B: \(\frac{6}{7}\)

Part (b)

\(\frac{6}{7}\)

Part (c)

\(P(A\cup B)\) (or \(\frac{6}{7}\))

Answer:

To solve this problem, we analyze each part step by step. First, we determine the number of favorable outcomes for each event and then calculate the probabilities. Finally, we use the principle of inclusion - exclusion for probability.

Part (a)

We assume that each outcome (numbered 1 - 7) is equally likely, so the total number of possible outcomes \(n = 7\).

Event A: The selected ball has a number from 2 to 5

The numbers from 2 to 5 are 2, 3, 4, 5. So the number of favorable outcomes \(n_A=4\).
The probability of event A, \(P(A)=\frac{n_A}{n}=\frac{4}{7}\)

Event B: The selected ball is grey

From the table, the grey - colored outcomes are 1, 4, 5, 7. So the number of favorable outcomes \(n_B = 4\).
The probability of event B, \(P(B)=\frac{n_B}{n}=\frac{4}{7}\)

Event A and B: The selected ball has a number from 2 to 5 and is grey

The numbers from 2 to 5 that are grey are 4, 5. So the number of favorable outcomes \(n_{A\cap B}=2\).
The probability of \(A\cap B\), \(P(A\cap B)=\frac{n_{A\cap B}}{n}=\frac{2}{7}\)

Event A or B: The selected ball has a number from 2 to 5 or is grey

Using the principle of inclusion - exclusion, \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). But we can also count the number of favorable outcomes directly. The outcomes in \(A\cup B\) are 1, 2, 3, 4, 5, 7. So \(n_{A\cup B}=6\).
The probability of \(A\cup B\), \(P(A\cup B)=\frac{n_{A\cup B}}{n}=\frac{6}{7}\)

Part (b)

We know that \(P(A)=\frac{4}{7}\), \(P(B)=\frac{4}{7}\) and \(P(A\cap B)=\frac{2}{7}\)
\[

$$\begin{align*} P(A)+P(B)-P(A\cap B)&=\frac{4}{7}+\frac{4}{7}-\frac{2}{7}\\ &=\frac{4 + 4-2}{7}\\ &=\frac{6}{7} \end{align*}$$

\]

Part (c)

From the principle of inclusion - exclusion in probability, we know that \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). From part (a), we found that \(P(A\cup B)=\frac{6}{7}\) and from part (b) we calculated \(P(A)+P(B)-P(A\cap B)=\frac{6}{7}\). So \(P(A)+P(B)-P(A\cap B)=P(A\cup B)\)

Final Answers

Part (a)

• Event A: \(\frac{4}{7}\)
• Event B: \(\frac{4}{7}\)
• Event A and B: \(\frac{2}{7}\)
• Event A or B: \(\frac{6}{7}\)

Part (b)

\(\frac{6}{7}\)

Part (c)

\(P(A\cup B)\) (or \(\frac{6}{7}\))