Hello everyone, I have the question below. I don't have the exact solution, so I don't know what the real answer is. Could you please explain the solution process step by step?

After that paper is drawn, there will be 41 papers from Section 2 among the 93 papers remaining. So, for the second paper drawn, the probability it is from section 2 is ...

41 / (41 +52) = 41/93

Similarly, for the 3rd and 4th papers drawn, the probability they are from Section 2 is, respectively,

40/92 . . . . 3rd paper

39/91 . . . . .4th paper

Then the probability that all 4 papers are from Section 2 is the product of these:

p(4 from section 2) = (42/94)(41/93)(40/92)(39/91) = 2,686,320/73,188,024

p(4 from section 2) ≈ 0.0367

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Alternate approach

There are 42C4 = 111,930 ways to choose 4 papers from 42, so this is the number of ways that 4 papers can be drawn from those written by Section 2 students.

There are 94C4 = 3,049,501 ways to choose 4 papers from the collection of 94 papers from both sections.

Hence the probability that all 4 are from Section 2 is ...

p(4 from section 2) = (ways to draw 4 from section 2)/(ways to draw 4)

= 111,930/3,049,501

p(4 from section 2) ≈ 0.0367

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Of course, you know that nCk means n!/(k!(n-k)!), where ! is the factorial symbol meaning the product of all positive integers to and including the one indicated.