Question 1192023: Hello, I have this problem I can't really solve: In the box there are balls with 1, 2, ..., 10 numbers on them. Two balls are randomly pulled one after the other with a return.
a) What is the probability that the sum of the numbers written on the balls will be greater than 14?
b) Then when pulling out the first ball, we see it with the even number on it. What is the probability that the sum of the numbers written on the balls will be greater than 14?
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! If a ball is returned, this is like two 10-side dice being rolled.
There are 100 different outcomes from 1/1 to 10/10
The smallest sum is 2, the largest 20, each of those two values with 1/100 probability
3 will have 2/100 probability (1/2 an 2/1)
4 will have 3/100...
from the other side
20 will have 1/100 probability
19 will have 2/100 (9/10. 10/9)
18 will have 3/100 (8/10, 10/8, 9/9)
17 will have 4/100
16 will have 5/100
15 will have 6/100 probability (7/8,8/7,6/9,9/6,5/10,10/5).
The answer is 21/100.
the most probable outcome is a sum of 11 with 6 and 5; 7 and 4; 8 and 3; 9 and 2; 10 and 1, each twice or 1/10 probability. Contrast this with two cubic die with a maximum sum of 12, the most probable one of 7 (1 more than the number of sides), with a probability of 1/6 (as if one die were rolled).
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if the ball has an even number, there are 5 possibilities, 2,4,6,8,10
Two of these possibilities can be excluded immediately, 2 and 4, since ball 2 won't make a difference.
if it is a 10, with probability 1/10, 6 of the 10 possibilities of ball 2 will work, so that probability is (1/10)(3/5)=3/50.
If it is an 8, 4 of 10 possibilities will work, so that probability is (1/10)(2/5)=2/50.
If it is a 6, 2 of 10 possibilities will work, so that probability is (1/10)(1/5)=1/50.
The answer is the sum of those 3 or 6/50 or 12%.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The number of possible outcomes is 10*10 = 100.
(a) P(sum greater than 14)
Outcomes with sum greater than 14:
1st ball 10: 2nd ball 5, 6, ..., 10 (6 possibilities)
1st ball 9: 2nd ball 6 to 10 (5 possibilities)
1st ball 8: 2nd ball 7 to 10 (4)
1st ball 7: 2nd ball 8 to 10 (3)
1st ball 6: 2nd ball 9 or 10 (2)
1st ball 5P 2nd ball 10 (1)
Total good outcomes: 6+5+4+3+2+1 = 21
P(sum greater than 14) = 21/100
(b) P(sum greater than 14, given that the first ball is even)
Using the list above, now the good outcomes are only if the first ball is 10, 8, or 6.
Number of good outcomes: 6+4+2=12
P(sum greater than 14) = 12/100
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