Question

Consider the trial on which a 3 is first observed in successive rolls of a six-sides die. Let A be the event that 3 is observed of the first trial. Let B be the event that at least 2trials are required to observe a 3. Assuming that each side has probability 1/6, find
1) P(A)
2) P(B)
3) P(A or B)
Please, help.

Answer 1

@phi

Answer 2

A is prob of rolling a 3
# of successes/ total # = 1/6

Answer 3

use binomial

Answer 4

event B = Pr # of rolls required > 1
which = 1 - Pr(1 roll) = 5/6

Answer 5

P(x>2) = 1 - P(x<2)

Answer 6

I know the answer, but I don't know how to put it in logic, first of, I have to define A, and B as sets, How? Please

Answer 7

Like A = {x | x...... something I don't know how to write}

Answer 8

I have never done this type of problem.
if A is a set, the most obvious is { 3 }
B, on the other hand is probably infinite.. consisting of e.g.{ 13,23,43,....}

Answer 9

but that hardly seems like a useful way to define probabilities.

Answer 10

Oh, I see
A= {3}
B={\(3^c3,3^c3^c3,3^c3^c3^c3,....\)}
Thank you @phi
I got it

Answer 11

so that P(A) =1/6
and P(B) =5/6