Question 418405: please help me solve this question on probability and statistics. Thank you...There is a jar containing 10 balls. The balls are marked with integers from 1 through 10. A ball is randomly withdrawn (with replacement) 10 times (x1, x2,..x10)from the jar and the number marked on it noted down. What is the probability that out of the 10 numbers noted down, the integer, 1, appears twice but not near each other, for example, (x, 1, x, x, 1, x, x, x, x, x) or (1, x, x, x, x, x, x, x, x, 1) or (x, x, 1, x, 1, x, x, x, x, x)...and so on.
Found 2 solutions by sudhanshu_kmr, stanbon:
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!
total no. of ways to withdrawn ball 10 times = 10*10..10times = 10^10
total no. of ways to put two 1's but not near = 36
no. of ways to withdrawn ball 10 times when two 1's appear but not near
= 36 * 9^8
(on another 8 places, any one of 2-10 can appear i.e except 1)
probability = (36 * 9^8) / (10^10)
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Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! please help me solve this question on probability and statistics. Thank you...
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There is a jar containing 10 balls. The balls are marked with integers from 1 through 10.
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A ball is randomly withdrawn (with replacement) 10 times (x1, x2,..x10)from the jar and the number marked on it noted down. What is the probability that out of the 10 numbers noted down, the integer, 1, appears twice but not near each other, for example, (x, 1, x, x, 1, x, x, x, x, x) or (1, x, x, x, x, x, x, x, x, 1) or (x, x, 1, x, 1, x, x, x, x, x)...and so on.
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On each draw there are 9 ways for the 1's to be next to one another.
There are 10C2 = 45 ways to place 2 one's in each draw.
So there are 45-9 = 36 ways for 2 one's to be separated.
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P(2 1's are not next to one another in a sequence of 10) = 36/45 = 4/5
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P(you have 2 1's) = ?
This is a binomial Problem with n = 10 trials and p(getting a 1) = 1/10
P(you have 2 1's ) = 10C2(1/10)^2*(9/10)^8 = 0.1937
That means there are 10C2 = 45 ways to place the 2 one's
The probability of 2 ones is (1/10)^2
The probability of 8 other digits is (9/10)^8
The probability you have 2 1's is 45*(1/10)^2*(9/10)^8 = 0.1937
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Not the answer but some of this might help.
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Cheers,
Stan H.
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