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?

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?
Transcript text: 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?
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Solution

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Solution Steps

Step 1: Probability of Selecting Two Strong Readers

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

P(X=k)=(Kk)(NKnk)(Nn) P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}

For strong readers, we have:

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

Calculating this gives:

P(X=2)=(32)(50)(82)=0.1071 P(X = 2) = \frac{\binom{3}{2} \binom{5}{0}}{\binom{8}{2}} = 0.1071

Step 2: Probability of Selecting Two Average Readers

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

For average readers, we have:

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

Calculating this gives:

P(X=2)=(32)(50)(82)=0.1071 P(X = 2) = \frac{\binom{3}{2} \binom{5}{0}}{\binom{8}{2}} = 0.1071

Step 3: Probability of Selecting Two Weak Readers

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

For weak readers, we have:

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

Calculating this gives:

P(X=2)=(22)(60)(82)=0.0357 P(X = 2) = \frac{\binom{2}{2} \binom{6}{0}}{\binom{8}{2}} = 0.0357

Step 4: Total Probability of Selecting the Same Type of Reader

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

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

Final Answer

The total probability that both selected students are of the same type of reader is:

0.2499 \boxed{0.2499}

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