larger than 9? that means 10,11, 12
Well, to get a 12, you can do it one way: Pr=1/6*1/6
To get an 11, you can do it two ways (5,6, or6,5) Pr= 2*1/6*1/6
To get a 10, you can do it 64,46,5,5 three ways
Pr=3*1/6*1/6
add the probabilities of all the combinations.
Roll two standard dice and add the numbers. What is the probability of getting a number larger that 9 for the first time on the third time.
5 answers
Actually, rereading the question, I don't know what "for the first tme on the third time" means.
I seem to get 0.167--book has 0.1157
Thank you for trying--appreciate it.
It means that the first two times you get a 9 or less, but the third time you get a 10, 11 or 12.
For third time, add the probabilities from Bobpursley. For either-or probabilities, add individual probabilities.
(1/6*1/6) + (2*1/6*1/6) + (3*1/6*1/6) = ?
Probability of 9 or less = 1 -?
For probability of all events, multiply probabilities of individual events. Thus the answer to your problem is:
(1-?)(1-?)(?) = ??
I'll let you do the calculations.
For third time, add the probabilities from Bobpursley. For either-or probabilities, add individual probabilities.
(1/6*1/6) + (2*1/6*1/6) + (3*1/6*1/6) = ?
Probability of 9 or less = 1 -?
For probability of all events, multiply probabilities of individual events. Thus the answer to your problem is:
(1-?)(1-?)(?) = ??
I'll let you do the calculations.