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Let's break down the problem step by step:

### a) State two assumptions Karen needs to make to use her model.

Karen is using a binomial model for the random variable \( X \), which represents the number of successes (catching the mouse) in 5 attempts with a success probability of 0.2. To use the binomial distribution model, Karen would need to assume:
1. Each attempt is independent of the others (the result of one attempt doesn't affect the others).
2. The probability of success (catching the mouse) on each attempt remains constant at 0.2.

### b) Using Karen’s model, find \( P(X \geq 3) \).

The random variable \( X \) follows a binomial distribution \( X \sim \text{Bin}(5, 0.2) \), where the number of trials is 5 and the probability of success on each trial is 0.2.

To find \( P(X \geq 3) \), we calculate:

\[
P(X \geq 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)
\]

The probability mass function of a binomial distribution is:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

So, let's compute these probabilities:

1. \( P(X = 0) = \binom{5}{0} (0.2)^0 (0.8)^5 = 0.32768 \)
2. \( P(X = 1) = \binom{5}{1} (0.2)^1 (0.8)^4 = 5 \times 0.2 \times 0.4096 = 0.4096 \)
3. \( P(X = 2) = \binom{5}{2} (0.2)^2 (0.8)^3 = 10 \times 0.04 \times 0.512 = 0.2048 \)

Adding these gives:

\[
P(X = 0) + P(X = 1) + P(X = 2) = 0.32768 + 0.4096 + 0.2048 = 0.94208
\]

Thus,

\[
P(X \geq 3) = 1 - 0.94208 = 0.05792
\]

### c) Using Karen’s assumptions about this experiment, find \( P(Y = 3) \).

Karen's assumptions imply that \( Y \), the number of attempts needed to catch the mouse for the first time, follows a geometric distribution with parameter \( p = 0.2 \). The probability of catching the mouse on the \( n \)-th attempt is given by:

\[
P(Y = n) = (1 - p)^{n-1} p
\]

For \( n = 3 \):

\[
P(Y = 3) = (0.8)^2 \times 0.2 = 0.64 \times 0.2 = 0.128
\]

### d) Find the value of \( k \).

Ian models \( P(Y = n) \) as:

\[
P(Y = n) = 0.1 + (n - 1)k
\]

Given that the probabilities must sum to 1 over \( n = 1, 2, 3, 4, 5 \) (since no cat will need more than 5 attempts), we can write:

\[
P(Y = 1) + P(Y = 2) + P(Y

In an experiment, a group of cats each repeatedly try to catch a toy mouse. For each cat, the random variable X represents the number of times the cat catches the mouse in the first five attempts. Karen models X as B(5, 0.2). State two assumptions Karen needs to make to use her model. Using Karen’s model, find P(X ≥ 3). For each cat, the random variable Y represents the number of attempts on which it first catches the mouse. Using Karen’s assumptions about this experiment, find P(Y = 3). Ian assumes that in this experiment, no cat will need more than 5 attempts to catch the mouse for the first time. He models P(Y = n) as P(Y = n) = 0.1 + (n – 1)k. Find the value of k. Using Ian’s model, find P(Y = 3). Explain how Karen’s and Ian’s models differ in describing the probability that a cat catches a mouse in this experiment.

This problem explores the probability analysis of cats catching a toy mouse using binomial and geometric distributions. Learn how to calculate P(X ≥ 3) using Karen's model, find the value of k in Ian's model, and understand the differences in their approaches. Suitable for students in Grades 10-12.

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