Binomial Distribution

6iii)

At least once means $X \geq 1$.

Exactly once means $X=1$.

So for part (iii) you seek $P(X \geq 1)$ ... which can either be done by computing $P(X=1) + \ldots + P(X=6)$, or taking the shortcut by considering the fact that all $P(0 \leq X \leq 6) = 1$.

Can you explain the short method i dont understand how to work out x from that

You're not supposed to 'work out X' it's a random variable which takes on values 0,1,...,6 and since it's binomially distributed you are able to compute the probabilities of it taking on any of these values.

You can easily work out what each of $P(X=1), \ldots, P(X=6)$ are and just add them up for $P(X \geq 1)$.

The reason I say there is a shortcut is because all probabilities add up to 1:

$P(X=0) + P(X=1) + \ldots + P(X=6) = 1$

and this is equivalent to

$P(X=0) + P(X \geq 1) = 1$

So just rearrange to get

$P(X \geq 1) = 1-P(X=0)$

I have another question this is to do woth significance level, i just dont know which values to assign n and p to ill attach my working and the question

2ii)

im not sure if ive done the the X~B part right so i stopped

Im sorry to keep asking but i dont know what i should set H(null) and H(one) to for this question and even if i did which year i would choose

6i)

There is my working although i think it is completely wrong

it gives me an answer above one

Check what you wrote down... 2/3 appears TWICE ??