For example, 1,2,3,4,5,6,1,5,3,3

I'm thinking the answer is as follows:

f(x)=P(X=x)=1/n for i=1,2,….,n

Ans: X=10=n=1/10

am i correct or absolutely incorrect heh ?

- Thread starter ITman95
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For example, 1,2,3,4,5,6,1,5,3,3

I'm thinking the answer is as follows:

f(x)=P(X=x)=1/n for i=1,2,….,n

Ans: X=10=n=1/10

am i correct or absolutely incorrect heh ?

I dont understand what you mean "if yes you should not in doubt" but the number after the asterisk is the probability of rolling any number on a die, 1/6 since there are 6 numbers on a die.

My question is why are we using 5/6 ? this is the probability that any OTHER different number is rolled right

My question is why are we using 5/6 ? this is the probability that any OTHER different number is rolled right

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5/6 is the probability of not matching the previous roll. (continue rolling)

This is basically a negative binomial distribution, except on the first roll probability of stopping = 0 and the probability of continuing is 1.

\({k + r -1}\choose{k} \)\( * (1 - p)^r * p^k\)

where

r= number of failures = 1

k \(\in\) {1,2,3 ... } = number of successes

p = probability of success = 5/6

(1 - p) = probability of failure = 1/6

By convention we usually call the stopping event (rolling the same number twice in a row) a failure. It may seem more natural to call this a success, but it just means the stopping event.

Because r=1 the first term simplifies to \({k + r -1}\choose{k}\) = \({k}\choose{k}\) = 1

So so for P(X = 10) we need

SSSSSSSSSF = \(1 * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{1}{6}\)

I think you made a mistake wouldnt p(x=1)=1/6 and p(x=2)=(5/6)*(1/6) and p(x=3)=(5/6)^2 * (1/6) and so on so that its P(X = x) = (5/6)(x-1) × (1/6) or is this just in the case where if you are looking to roll a specific number for example a 5 instead of a repeat which is whats asked in the question ?

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