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In a class of 17 students, 3 are math majors. A group of four students is chosen at random.

(a) What is the probability that the group has no math majors?
(b) What is the probability that the group has at least one math major?
(c) What is the probability that the group has exactly two math majors?

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Answer:

Explanation:

Step1

Definition permutation (order is important):

P(n,r)= n!/(n−r)!

Definition combination (order is not important):

C(n,r) =( n/r )= n!/r!(n−r)!

Step2

a) The order of the students does not matter, because a different order results in the same students being selected, and thus we should use the definition of combination.

We are selecting 3 of the 17 students in the class (of which 3 are math majors and 14 are not math majors).

# of possible outcomes=(17,3)=17!/3!(17−3)!=17!/3!14!=680

There areC(14,3) ways to select 3 of the 14 no math majors.

# of favorable outcomes=C(14,3)= 14!/3!(14−3)! = 14!/3!11! =364

The probability is the number of favorable outcomes divided by the number of possible outcome.

P(no math majors)= # of favorable outcomes/# of possible outcomes

= 364/680= 91/170

≈0.5353=53.53%

Step 3

(b) By part (a):

P(no math majors)=91/170≈0.5353=53.53%

Use the Complement rule

Complement rule:

P(E’)=1−P(E)

P(at least one math major)=1−P(no math majors)=1−91/170

=79/170≈0.4647=46.47%

Step 4

(c) The order of the students does not matter, because a different order results in the same students being selected, and thus we should use the definition of combination.

We are selecting 3 of the 17 students in the class (of which 3 are math majors and 14 are not math majors).

# of possible outcomes=C(17,3)= 17!/3!(17−3)!=17! /3!14!=680

There areC(14,1) ways to select 1 of the 14 no math majors and there are

C(3,2) ways to select 2 of the 3 major majors.

# of favorable outcomes=C(14,1)⋅C(3,2)=14⋅3=42

P(2 math majors)= # of favorable outcomes/# of possible outcomes

= 42/680= 21/340=approx0.0618=6.18%

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