Question

six sides of cube are marked as follows: three sides with number -3; two sides with number 2; and one side with number 0.

a) what is the probability that in 4 tries maximum once you get number -3 (i solved this one and got 1/16)

b) what is the probability that in 3 throws you get that sum of numbers is negative

c) how many times you must throw dice so that you get 0 once with 0,999 probability

d) cube is repeatedly thrown with rules:
- if you get negative number nothing happens
- if you get 0 or 2, you replace that number with -3
you throw so long until all sides of cube

Answer 1

d) ...you throw so long until all sides of cube are -3. what is the probability that it would take more than 3 throws to achieve that.

this is for my friend, its been long time i did tasks like this, so i hoped you could help...
tnx

Answer 2

a- 15/16. probability of that you don't get number -3 is 1/2. if you don't get number -3 four times, (1/2)*(1/2)*(1/2)*(1/2)=1/16.
1-(1/16)=15/16

Answer 3

yes but probability that you only once get -3...
probability to get -3 = 1/2
probability not to get -3 = 1/2
so:
(1/2)^4=1/16