Hi
VeritasKarishmaIf a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?
A. 25/216
B. 50/216
C. 25/72
D. 25/36
E. 5/6
Total ways in which a 6-sided die can be rolled three times = 6*6*6 = 216
To get exactly one 3, there are three ways:
A 3 on the first roll and non 3 on other two rolls. This can be done in 1*5*5 = 25 ways.
The 3 could be on the second or third roll too. So total favorable cases = 25*3 = 75
Required Probability = 75/216 = 25/72
This is one of your solutions for the question.
I request your advice on this question and probability in general.
SO in this question:
I went about it in this way:
Total cases: 216
Desired:
3AB: (If A and B are different)
total cases: 5 *4=20
now since all numbers are different. the total number of combinations possible is: 3! *20= 120
3AA (if a is same): A can be 1,2,4,5,6 Combinations possible: 3!/2!=3 i.e 15
Total cases: 135
which is wrong.
I dont know why/where I am going wrong.
Second problem:
I have been scoring above Q50 on all
veritas mocks but I am really struggling with probability. There are some pretty basic stuff that confuse me (such as the question above). I have the most problems when I have to choose and order is important. In those cases, I am generally wrong.
Where can I work on these specific problem?
Is there any resource/article I can use?
Looking forward to your reply.
Regards
Nitesh