we roll a dice multiple times. Let X be a random variable for the number of rolls until the first 2 occurs.
we can say that X is geometric distributed with density function f(k) = p(1-p)^k-1
where p = 1/6
and k € N
a) calculate P(X<3)
b) calculate the probability that it takes more than 2 rolls to get the first 2.
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Since `f(k)=p(1-p)^(k-1)` is probability density function of geometric random variable it follows that its cumulative distribution function is
`F(k)=sum_(i=1)^k f(i)=sum_(i=1)^k p(1-p)^(i-1)=p+p(1-p)+cdots+p(1-p)^(k-1)=`
Alternatively you can calculate `P(X<3)` as follows:
Probability that it takes more than 2 rolls to get first 2 is `P(X geq 3)` which is opposite event to getting first 2 in less than 3 rolls. Hence ` `
`P(X geq 3)=1-P(X<3)=1-11/36=25/36`.
Alternatively you can calculate it by using cumulative probability distribution function:
`P(X geq 3)= F(infty)-F(2)=1-11/36=25/36`
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