Please help with binomial distribution question. How did they arrive at these answers? The questions are attached.

Question

anonymous · Answer

attachment

ganeshie8 · Answer

still working on this ?

anonymous · Answer

Yes I am

anonymous · Answer

for part a) I thought it would be $$\left(\begin{matrix}4 \ 1\end{matrix}\right) * 0.7 * 0.3^3$$

anonymous · Answer

but it is not, it is actually 0.7* 0.3^3

anonymous · Answer

Sorry i meant 0.8 and 0.2...

kirbykirby · Answer

These look like geometric distribution problems, not binomial. The geometric distribution looks at the number of trials until the 1st success occurs.

anonymous · Answer

that means i have can 4 trials

kirbykirby · Answer

Yes. And for the geometric distribution, we have that if X is the number of trials until the 1st success, then
$$P(X=x)=(1-p)^{x-1}p, ~~~\text{for }x=1,2,3,... $$

anonymous · Answer

wouldn't p be raised so an exponent?

anonymous · Answer

it remains at one because of one successful event?

kirbykirby · Answer

yes indeed. There would be an exponent on the p if we talked about the negative binomial (which is a generalization of the geometric). Although, there would also be a "choose" factor as well

kirbykirby · Answer

There are $x-1$ trials that are failures (so they show up with probability $1-p$), and the x-th trial is the success, which occurs with probability p

anonymous · Answer

Thanks :)

kirbykirby · Answer

your welcome =]