Question

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 $8$? I WILL GIVE BRAINLIEST TO WHOEVER ANSWES ALL OF THEM CORRECT

Answer 1

Answer:

Step-by-step explanation:

1 ) No of total slips after addition = 11

8 slip with 3 on it

3 slips with 9  on it

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

=  (8 / 11)  x 3 + (3 / 11)  x 9

24 / 11 + 27 / 11 = 4.636

2 )

No of total slips after addition = 12

8 slip with 3 on it

4 slips with 9  on it

expectation value = probability of 3 x 3 + 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

expectation value = probability of 3 x 3 + 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 =

n = 38

Minimum of 38 has to be added .