Statistics Question Urgent Help

bump

Insufficient information to resolve the problem - unless it's to be left in terms of a parameter.

Please link the original question.

sorry bout that forgot to add this bit Given P(X=2) = 0.35
updated it at the top as well

From the question, then:

P(X=2) = P(X=4) = 0.35

P(X=1) = P(X=3) = ?

And use the fact that the total probability must be 1, to work out the unknown value.

From the question, then:

P(X=2) = P(X=4) = 0.35

P(X=1) = P(X=3) = ?

And use the fact that the total probability must be 1, to work out the unknown value.

P(X=2) = P(X=4) = 0.35 how do you know 2 and 4 are related

Question tells you.

"P(X=q) = P(X=q+2) for q=1.2"

Setting q=2

P(X=2) = P(X=4)

so P(X=1) = 0.15
P(X=3) =0.15

Yes.

Can you do part b? If not, what distribution do you thnk it is?

Looks like binomial distribution so I've done
1-P(X<25) = P(X>25)
1-P(X{less or equal} 25) = P(X>25)

let x be 0.35

0.989956 for P(X<25)
1-0.989956=0.01
not too sure on this bit if I've done It right

Yes, should be $\leq$



p=0.35
Probability of success on a single spin.



Yep.

y=12/X
FindP(Y−X≤4)
I was thinking about finding in the 50 spins where X is less or equal to 4
But how would I do the Y bit?
Sorry about no latex I can't figure out how to use it

The 50 spins don't come into it, unless there's something you've not posted.

You're only dealing with 1 spin, with r.v. X

So, you just need to check when $(12/X)- X \leq 4$. There are only 4 values, with their corresponding probabilities.

If P(X<=4) then
all values less than it is 0.65 and including is equal to 1
cuz P(X=4) is 0.35 all values less is 0.65

OK, I think I know what you mean, but so what!

You're interested in P(12/X-X<= 4), not P(X<= 4)

You need to evaluate that function over the domain of X, i.e. 1,2,3,4. And check if that function is <=4.

I don't know where to start