To answer this item, we are to use the concept of combination. The equation is as follows,
nCr = (n!)/((r!)(n - r))!
Substituting the known values in the equation,
3C2 = 3!/(2!)(3-2)! = 3
Hence, the answer to this item is 3.
<em>ANSWER: 3</em>
Answer:
.
Step-by-step explanation:
We have been given two sets as A: {71,73,79,83,87} B:{57,59,61,67}. We are asked to find the probability that both numbers are prime, if one number is selected at random from set A, and one number is selected at random from set B.
We can see that in set A, there is only one non-prime number that is 87 as it is divisible by 3.
So there are 4 prime number in set A and total numbers are 5.

We can see that in set B, there is only one non-prime number that is 57 as it is divisible by 3.
So there are 3 prime number in set B and total numbers are 4.

Now, we will multiply both probabilities to find the probability that both numbers are prime. We are multiplying probabilities because both events are independent.



Therefore, the probability that both numbers are prime would be
.
There is no question mark in the question
and the statement is true