Question

7 fill in the blank 8 points if there was only one government class, and wes and maureen were in that class of 38 students, what would be the probability that both wes and maureen would be selected as the two students to go to washington? is this still an example of independent events? choose your answer...

Explanation:

Step1: Calculate total ways to choose 2

The total number of ways to choose 2 students from 38 is given by the combination formula \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 38 \) and \( k=2 \). So \( C(38,2)=\frac{38!}{2!(38 - 2)!}=\frac{38\times37}{2\times1}=703 \).

Step2: Calculate favorable ways

There is only 1 favorable way (choosing Wes and Maureen together).

Step3: Find probability

The probability \( P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{1}{703} \approx 0.00142 \).

Step4: Check independent events

For independent events, \( P(A\cap B)=P(A)\times P(B) \). Let \( A \) be selecting Wes and \( B \) be selecting Maureen. \( P(A)=\frac{2}{38} \), \( P(B)=\frac{2}{38} \) (if we consider selection with replacement, but here it's without replacement). \( P(A)\times P(B)=\frac{2}{38}\times\frac{2}{38}=\frac{4}{1444}
eq\frac{1}{703} \). Also, the selection of one affects the probability of the other (since after selecting one, the number of students left changes), so they are dependent events.

Answer:

The probability is \(\frac{1}{703}\) (or approximately \(0.00142\)) and this is not an example of independent events (the two events are dependent).