Answer:
7/15
Explanation:
Given 3 of 10 counters are blue, you want the probability that randomly drawing 2 without replacement will result in drawing exactly one blue counter.
Probability
The probability of drawing a counter that is blue is the ratio of the number of blue counters to the total number of counters. The probability of drawing one that is not blue is the same: the ratio of not-blue counters to the number of counters.
The probability of drawing a particular sequence of counters is the product of the probabilities of the members of that sequence.
One blue
There are two ways to get exactly one blue:
- blue on the first draw and not-blue on the second
- not-blue on the first draw and blue on the second
The respective probabilities of these events are ...
(blue, not-blue) = (3/10)×(7/9) = 7/30
(not-blue, blue) = (7/10)×(3/9) = 7/30
The probability of one or the other of these is the sum of their probabilities:
p(only one blue) = 7/30 +7/30
p(only one blue) = 7/15
__
Additional comment
There are other ways this can be computed.
For example, there are 3·7 = 21 ways to have one blue and one not-blue counter. There are 10·9/(2·1) = 45 ways to draw 2 counters out of 10. So the probability of one blue in a draw of two is 21/45 = 7/15.