If two letters are chosen at random (without replacement) from the word JUPITER, what is the probability that none of the letters are vowels?	Choices:
A) 1/7
B) 2/7
C) 3/14
D) 5/14
E) 11/21
The number of ways to choose two letters is 7C2 = 21   The number of ways to choose two consonants is 4C2 = 6.   P(no vowels) = 6/21 = 2/7   The answer is 2/7.
<result>B</result>
If two letters are chosen at random (without replacement) from the word JUPITER, what is the probability that none of the letters are vowels?