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steposvetlana [31]
2 years ago
14

Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $3$ on them and the other two have a $9$ on them. How m

any $9$'s do we have to add to make the expected value equal to $6$
Mathematics
1 answer:
8_murik_8 [283]2 years ago
4 0

Value of the given statement:

  1. 4.36
  2. 5
  3. 6
  4. 38
<h3>What is Probability?</h3>

The area of mathematics known as probability deals with numerical descriptions of how likely it is for an event to happen or for a claim to be true. A number between 0 and 1 is the probability of an event, where, broadly speaking, 0 denotes the event's impossibility and 1 denotes its certainty.

According to the given information:

1) There are a total of 10 slips after adding.

8 slip with 3 on it

2 slips with 9  on it

 value = Probability of 3 x 3 plus Probability of 9 x 9.

=  (8 / 11)  x 3 + (2/ 11)  x 9

24 / 11 + 24 / 11 = 4.36

2 )No of total slips after addition = 12

8 slip with 3 on it

4 slips with 9  on it

 value = Probability of 3 x 3 plus Probability of 9 x 9.

=  (8 / 12)  x 3 + (4 / 12)  x 9

2 + 3 = 5

3 )Let n be the required number

No of total slips after addition = 10+n

8 slip with 3 on it

2 + n  slips with 9  on it

value = Probability of 3 x 3 plus Probability of 9 x 9.

=  (8 / 10+n )  x 3 + (2+n  / 10+n )  x 9 = 6

24 + 18 + 9n / 10 + n  = 6

42 + 9n = 60 + 6n

3 n = 18

n = 6

4 )Let n be the required number

No of total slips after addition = 10+n

8 slip with 3 on it

2 + n  slips with 9  on it

expectation value = probability of 3 x 3 + probability of 9 x 9

=  (8 / 10+n )  x 3 + (2+n  / 10+n )  x 9 = 8

24 + 18 + 9n / 10 + n  = 8

42 + 9n = 80 + 8n

n = 38

Minimum of 38 has to be added .

To know more about Probability visit:

brainly.com/question/17487084

#SPJ4

I understand that the question you are looking for is:

1:Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $3$ on them and the other two have a $9$ on them. What is the expected value of the number shown if we add one additional $9$ to the bag? 2:Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $3$ on them and the other two have a $9$ on them. What is the expected value of the number shown if we add two additional $9$'s (instead of just one) to the bag? 3:Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $3$ on them and the other two have a $9$ on them. How many $9$'s do we have to add to make the expected value equal to $6$? 4:Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $3$ on them and the other two have a $9$ on them. How many $9$'s do I have to add before the expected value is at least $8$?

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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
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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)

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Step-by-step explanation:

to find C we have to plug in the coordinate given to us

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