Question

a

Answer 1

we have a general formula for this type of distribution
\[P(X=x) = p^x*(1-p)^{1-x}\]

Answer 2

@niksva, it should be
\[P(X=x)=p(1-p)^{x-1}\]

Answer 3

X is a random variable
while x is no. of trials
in the first part
\[P(X>=2) = 1 - P(X<2)\]
\[P(X>=2)= 1-P(X=0)-P(X=1)\]

Answer 4

yeah @SithsAndGiggles is right

Answer 5

\(X\) follows a geometric distribution, which gives you the probability of the first success occurring after a number of failures.

\(P(X=x)\) can be interpreted as the probability of success on the \(x\)-th trial, following \(x-1\) failures.

Answer 6

Yes, \(p\) is given to be \(0.1\).

And yes, that's what \(P(X\ge2)\) basically means.

Answer 7

Sure

Answer 8

\[\begin{align*}P(X\ge2)&=1-P(X<2)\\
&=1-P(X=1)&\text{assuming the support of }X\text{ is positive}\\
&=1-p(1-p)^{1-1}\\
&=1-0.1(0.9)^0\\
&=0.9
\end{align*}\]

Answer 9

Yes, think of it as "the minimum of trials must be 1."

Answer 10

pleasure
@SithsAndGiggles explained it really well