6) It is the last week of the football season. In order for the Knights to make the playoffs, the Knights must win and the Cougars must lose. The Knights and the Cougars do not play against each other. P (Knights win) = 0.75 P (Cougars lose) = 0.40 a) Are the events "Knights Win" and "Cougars Lose" independent? Justify your answer. b) What is the probability that the Knights will make the playoffs? Show how you determined your answer. 7) There are three boys in a class of nine students. Three students are to be randomly selected (without replacement) to help with bus duty. Determine the probability that all three students are boys. Show how you determined your answer.

Sure, let's tackle each question step by step.

6)

a) Are the events "Knights Win" and "Cougars Lose" independent? Justify your answer.

Two events are independent if the occurrence of one event does not affect the probability of the other event occurring. Mathematically, events A and B are independent if:

$P(A \cap B) = P(A) \times P(B)$

In this case:

• Event A: Knights win
• Event B: Cougars lose

We need to check if:

$P(\text{Knights win and Cougars lose}) = P(\text{Knights win}) \times P(\text{Cougars lose})$

Given: $P(\text{Knights win}) = 0.75$ $P(\text{Cougars lose}) = 0.40$

Assuming the events are independent: $P(\text{Knights win and Cougars lose}) = 0.75 \times 0.40 = 0.30$

Since we are not given any information that suggests otherwise, we assume the events are independent. Therefore, the events "Knights Win" and "Cougars Lose" are independent.

b) What is the probability that the Knights will make the playoffs? Show how you determined your answer.

The Knights will make the playoffs if both the Knights win and the Cougars lose. Since we have already assumed these events are independent, we can multiply their probabilities:

$P(\text{Knights make playoffs}) = P(\text{Knights win}) \times P(\text{Cougars lose})$

$P(\text{Knights make playoffs}) = 0.75 \times 0.40 = 0.30$

So, the probability that the Knights will make the playoffs is 0.30.

7)

There are three boys in a class of nine students. Three students are to be randomly selected (without replacement) to help with bus duty. Determine the probability that all three students are boys. Show how you determined your answer.

First, we need to determine the total number of ways to choose 3 students out of 9. This can be calculated using the combination formula:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Where $n$ is the total number of students, and $k$ is the number of students to be chosen.

$\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$

Next, we need to determine the number of ways to choose 3 boys out of the 3 boys available:

$\binom{3}{3} = \frac{3!}{3!(3-3)!} = \frac{3!}{3!0!} = 1$

The probability that all three students selected are boys is the ratio of the number of favorable outcomes to the total number of outcomes:

$P(\text{all three students are boys}) = \frac{\binom{3}{3}}{\binom{9}{3}} = \frac{1}{84}$

So, the probability that all three students selected are boys is $\frac{1}{84}$.