Questions: There are 8 students in a reading group. Three of the students are classified as strong readers, three as average and two as weak readers. A researcher wants to work with 2 randomly selected students from this group. What is the probability that both of the students she selects are the same type of reader?

Solution Steps

To find the probability that both selected students are strong readers, we use the hypergeometric distribution formula:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

For strong readers, we have:

• \(N = 8\) (total students)
• \(K = 3\) (strong readers)
• \(n = 2\) (students selected)
• \(k = 2\) (strong readers selected)

Calculating this gives:

\[ P(X = 2) = \frac{\binom{3}{2} \binom{5}{0}}{\binom{8}{2}} = 0.1071 \]

Next, we calculate the probability that both selected students are average readers using the same formula:

For average readers, we have:

• \(N = 8\)
• \(K = 3\) (average readers)
• \(n = 2\)
• \(k = 2\)

Calculating this gives:

\[ P(X = 2) = \frac{\binom{3}{2} \binom{5}{0}}{\binom{8}{2}} = 0.1071 \]

Now, we find the probability that both selected students are weak readers:

For weak readers, we have:

• \(N = 8\)
• \(K = 2\) (weak readers)
• \(n = 2\)
• \(k = 2\)

Calculating this gives:

\[ P(X = 2) = \frac{\binom{2}{2} \binom{6}{0}}{\binom{8}{2}} = 0.0357 \]

Finally, we sum the probabilities of selecting two students of the same type:

\[ P(\text{same type}) = P(\text{strong}) + P(\text{average}) + P(\text{weak}) = 0.1071 + 0.1071 + 0.0357 = 0.2499 \]

Final Answer