grenico wrote:
Asif123 wrote:
Two fair dice with sides numbered 1 to 6 are tossed. What is the probability that the sum of the exposed faces on the dice is a prime number?
A) 5/11
B) 5/12
C) 5/21
D) 2/9
E) 5/36
The list of possible sums of the two dice are:
2,3,4,5,6,7,8,9,10,11,12
From this list, 5 are prime:
2,3,5,7,11
Therefore, the probability is \(\frac{5}{11}\)
The answer is AThe problem here is that you are giving each of the possible sums an equal weight. That is, you are giving them an equal probability of occurring which is not the case.
The probability of getting a sum of 7 for example is (6/36), whereas the probability of getting a sum of 2 is (1/36).
So you don't have a simple sample space.
If you want to do it this way you have to first find the probability of the possible sums.
P(Sum=2)=1/36
P(Sum=3)=2/36
P(Sum=5)=4/36
P(Sum=7)=6/36
P(Sum=11)=2/36
Then we are looking for P(Sum=2 or Sum=3 or Sum=5 or Sum=7 or Sum=11)
Since each event is disjoint
P(Sum=2 or Sum=3 or Sum=5 or Sum=7 or Sum=11)
=P(Sum=2)+ P(Sum=3)+ P(Sum=5)+ P(Sum=7)+ P(Sum=11)
= (1/36)+ (2/36)+ (4/36)+ (6/36)+ (2/36)
= 15/36
= 5/12
Final Answer: B