1. Suppose that John and Tom are sitting in a classroom containing 9 students in total....

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math Statistics-And-Probability

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Answer 1

Answer #1

Number of ways of selecting 4 students out of 9 for group I and 5 students for group II is

9! C(9.4)C(5,5)=415! 5! 5!(5-5)! 1 26

(a)

Since we need John in first group so we need to select 3 more people for first group. So number of ways of selecting John in first group is

8! 5! C8,3)c(6,5) -315 5-5 = 56

The required probability is

P(John is first group)6 56 126 0.4444

(b)

Number of groups with John is first group is 56.

Number of groups with John and Tom in first group is

7! 5! = 21

So the required probability is

P(Tom is first group and John is first group P(John is first group) P(Tom is first group!John is first group) = 21 56 = 0.375

(c)

Number of groups with John and Tom in first group is

7! 5! = 21

Number of groups with John and Tom in second group is

7! 3! 35 C(7.4)C(3.3) _ 4!31 ·31(3-3)!

So the number of groups such that John and Tom are in the same group is

21 + 35 = 56

The required probability is

56 P(John and Tom in same group)- 126 0.4444

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