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loris [4]
2 years ago
12

A group of three undergraduate and five graduate students are available to fill certain student government posts. If four studen

ts are to be randomly selected from this group, find the probability that exactly two undergraduates will be among the four chosen.
Mathematics
1 answer:
creativ13 [48]2 years ago
5 0

Answer:

Pr = 0.4286

Step-by-step explanation:

Given

Let

U \to\\ Undergraduates

G \to Graduates

So, we have:

U = 3; G =5 -- Total students

r = 4 --- students to select

Required

P(U =2)

From the question, we understand that 2 undergraduates are to be selected; This means that 2 graduates are to be selected.

First, we calculate the total possible selection (using combination)

^nC_r = \frac{n!}{(n-r)!r!}

So, we have:

Total = ^{U + G}C_r

Total = ^{3 + 5}C_4

Total = ^8C_4

Total = \frac{8!}{(8-4)!4!}

Total = \frac{8!}{4!4!}

Using a calculator, we have:

Total = 70

The number of ways of selecting 2 from 3 undergraduates is:

U = ^3C_2

U = \frac{3!}{(3-2)!2!}

U = \frac{3!}{1!2!}

U = 3

The number of ways of selecting 2 from 5 graduates is:

G = ^5C_2

G = \frac{5!}{(5-2)!2!}

G = \frac{5!}{3!2!}

G =10

So, the probability is:

Pr = \frac{G * U}{Total}

Pr = \frac{10*3}{70}

Pr = \frac{30}{70}

Pr = 0.4286

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Eva8 [605]

Answer:

4. 21

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Step-by-step explanation:

(the answer for #4

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6 0
3 years ago
Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can
Thepotemich [5.8K]

Answer:

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)}

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)}

Step-by-step explanation:

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)} (second die is even)

B = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)} (sum of the two numbers is even)

C = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5)} (at least one in the pair is odd i.e one of the pair is odd or both are odd)

A bar = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (the pairs that are not in A)

B bar = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5)} (the pairs that are not in B)

C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (the pairs that are not in C)

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (intersection: the pairs that are common to both A and B)

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)} (intersection: the pairs that are common to both A and B bar)

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} (union: all the pairs in A bar and B )

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (intersection: the pairs that are common to both A bar and C)

6 0
3 years ago
A student solved this problem and said the answer is quarts
kenny6666 [7]
B. no the answer isn't reasonable, it should be about 3 quarts.
3 0
3 years ago
Please help me with this
kaheart [24]

Answer:

m9 = 115

m4 =110

m10=65

m11= 70

m8= 70

m5 = 65

m3 = 70

m14 = 115

4 0
3 years ago
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