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peripatein
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Hi,
I would like to confirm my answer to the following question in Statistics. Hopefully one of you will be willing to provide some feedback.
A die is tossed repeatedly, until a result (1-6) equal to one of the preceding results is obtained. For instance, (3,2,4,2) is a series of tosses halted at the fourth toss (as only after that toss has a number recurred). X denotes the number of tosses till 'success' is obtained. The question asks for the probability function of X.
I wasn't sure how to come up with a general formula for P(X), but I did manage to derive the following, which, hopefully, is correct:
P(X = 2) = 1/6
P(X = 3) = (5/6)(2/6) = 5/18
P(X = 4) = (5/6)(4/6)(3/6) = 5/18
P(X = 5) = (5/6)(4/6)(3/6)(4/6) = 5/27
P(X = 6) = (5/6)(4/6)(3/6)(2/6)(5/6) = 25/324
P(X = 7) = 5! / 6^5 = 5/324
Is it correct? How may I come up with a general formula for P(X)? I'd appreciate any comments and your assistance.
I would like to confirm my answer to the following question in Statistics. Hopefully one of you will be willing to provide some feedback.
Homework Statement
A die is tossed repeatedly, until a result (1-6) equal to one of the preceding results is obtained. For instance, (3,2,4,2) is a series of tosses halted at the fourth toss (as only after that toss has a number recurred). X denotes the number of tosses till 'success' is obtained. The question asks for the probability function of X.
Homework Equations
The Attempt at a Solution
I wasn't sure how to come up with a general formula for P(X), but I did manage to derive the following, which, hopefully, is correct:
P(X = 2) = 1/6
P(X = 3) = (5/6)(2/6) = 5/18
P(X = 4) = (5/6)(4/6)(3/6) = 5/18
P(X = 5) = (5/6)(4/6)(3/6)(4/6) = 5/27
P(X = 6) = (5/6)(4/6)(3/6)(2/6)(5/6) = 25/324
P(X = 7) = 5! / 6^5 = 5/324
Is it correct? How may I come up with a general formula for P(X)? I'd appreciate any comments and your assistance.
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