Question 442709: If 6 numbers are drawn promiscuously from the numbers 1 to 49 inclusive,how do we find the probability of the selection containing exactly one run of three consecutive numbers.
E G 123 8 28 38,
or
4 9 15 16 17 49.
My book,Random Variables,D Stirzaker, offers a solution but it is clear that the author has made a mistake .morleyken@rocketmail.com.Thank you.
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! When drawing your first number, you have a 1/49 chance of drawing any number.
If you draw a 1: You have a 1/48 chance of drawing a number consecutive to it.
If you draw a 49: You have a 1/48 chance of drawing a number consecutive to it.
If you draw anything else: You have a 1/24 chance of drawing a number consecutive to it.
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2nd draw:
If you have drawn a 1, and then a 2, you have a 1/47 chance of drawing a number consecutive to it.
If you have drawn a 49, and then a 48, you have a 1/47 chance of drawing a number consecutive to it.
If you have draw something other than that, then you have a 4/47 chance of drawing a number consecutive to it.
For instance you have drawn 11,17
10, 12, 16, 18 are all candidates.
We also excluded ... so
4/44 = 1/11 chance.
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On our third draw:
probability of 1,2,3 = (1/49)(1/48)(1/47)
probability of 49,48,47 = (1/49)(1/48)(1/47)
Anything else: 6/46 = (3/23)
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We now have one run.
probability of 1,2,3 and then no run = (1/49)(1/48)(1/47)(45/46)(39/45)(35/44)
Same probability with 49,48,47 and then no run
Our anything else probability is: (1/49)(1/24)(4/47)(3/23)(37/45)(33/44)
probability(B) = (61465/10068347520)
probability(C) = (61465/10068347520)
probability(A) = (14652/2517086880)
P(A or B or C) = 2* (61465/100068347520) + 14652/2517086880)
= 2* (61465/100068347520) + (14652/2517086880)
= (18682943161 / 2650260149893440)
Hopefully... lol
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