5 students are selected at random from a group of 12.

Question

A. Find the probability of Erik being selected

B. Probability of Erik and Jens are both selected

Answer 1

I see both of these questions as hypergeometric probability questions.

Mark M. hinted at the answer in his comment above.

Question 1: ₁₁C₄ / ₁₂C₅

Question2: ₁₀C₄ / ₁₂C₅

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Answer 2

How many groups of 5 could Erik be on (presuming there is only one Erik)?

How many groups of 5 ( with or without Erik) are there?

We can presume that order does not matter; even if it did, we would get the same answer.

Erik can be on ₁₁C₄ different groups. This is

11! / ( 4! × (11-4)! ) = 11×10×9×8 / (4×3×2×1) = 330

The total number of groups of 5 out of 12 choices is

12! / ( 5! × (12 - 5)! ) = 12 × 11 × 10 × 9 × 8 / ( 5 × 4 × 3 × 2 × 1 ) = 792

330 / 792 = 5 / 12

Note: If the order mattered, we would remove 7! from both of the counts above. That would lead to the same final answer.

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Response to comment:

Consider any group that Erik is on. There are four other members of that group. Since Erik is already in that group, there are 11 × 10 × 9 × 8 ways of filling out that group. That's 11! / 7!. We can remove the permutations from the numeratator by dividing by 4!, and by 5! for the denominator That is written as 11C4 for the numerator, and the corresponding denominator would then be 12C5.