A level Math

In a game a player pays an entrance fee of $n. He then selects
one number from 1, 2, 3 or 4 and rolls three 4-sided standard dice.
If his chosen number appears on all three dice he wins four
times his entrance fee.
If his number appears on exactly two of the dice he wins three
times the entrance fee.

If his number appears on exactly one dice he wins $1.
If his number does not appear on any of the dice he wins
nothing.

Profit ($) −n 1-n 2n 3n
Probability 27/64 ? ? ?

how do I find the probability? the first one I was thinking of like (3/4)^3 but i dont understand why the 1-n is also 27/64
please help, thanks

Its a binomial distribution with n=3 (n is hte number of experiments, not entrance fee) and p=1/4,q=3/4. So just do the usual binomial terms whch should answer your question.
Obviously you could draw it as a tree as well and think how many leaves there are for the 1-n case. Its equivalent to the binomial.

Hey dude, I would really love if you elaborate more on your answer, I have been stuck with this problem for over a week.

No reason you should be stuck for that long, but as a new user its worth reading the sticky at the top of the forum about posting what youve done / better describe what youre stuck with. You could sketch/think about a simple tree if youre confused, but you should have done a binomial distribution and the probabilities are given directly, as in #1.

right, do you remember that you had replied to this person? for the same question?

Yes, see #1 for their reply and #3 for yours.

Well, i think the question is not correct. because in the table - users profit is given as "2n", but nowhere is there any conditions given when the players earns a profit twice his entrance fees

Well, i think the question is not correct. because in the table - users profit is given as "2n", but nowhere is there any conditions given when the players earns a profit twice his entrance fees

Thanks a lot, for clearing it up... and by the way can you explain why I should keep E (X) as 0
it confuses me a bit

Is that a different question part or ...? The expected value of a random variable is simply its mean.

the same question, I don't understand why I should keep E (x) = 0, when I calculate the minimum charge/fees

Is that a different question part or ...? The expected value of a random variable is simply its mean.

And, also can you assist me in calculating the answer out, I got 1.28 something and so on, I am not sure if this in cents or dollars, and it is said I have to round it up to whole number - so how?

well, look at this, (-n)*(27/64) + (n-1)(27/64) + (2n)(9/64) + 3n(1/64) = (21n - 27)/64 = 0, so 21n = 27, n = 27/21 which is nearly 1.28 something

Ahh, it is very late here, i am making so many mistakes - thanks, I got 0.81818181818181818181818181818182, next do I round it to 0.82 then write is 82 cents? or is is it supposed be just $ 0.82