A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
A. 21/50
B. 3/13
C. 47/50
D. 14/15
E. 1/5
I am getting a different answer when I solve it in the following way:
Probability of getting atleast one marble blue = Prob of selecting 2 blue marbles + Prob of selecting 1 blue marble and 1 red marble
i.e. (7/10)*(6/9) + (7/10) * (3/9) = 21/30 = 7/10
But the answer is D 14/15.
This is got by 1 - probability of picking both the marbles as red
i.e. (3/10) * (2/9) = 1/15
Therefore 1 - 1/15 = 14/15
Can anybody explain why the answers differ. Thanks.
A. 21/50
B. 3/13
C. 47/50
D. 14/15
E. 1/5
I am getting a different answer when I solve it in the following way:
Probability of getting atleast one marble blue = Prob of selecting 2 blue marbles + Prob of selecting 1 blue marble and 1 red marble
i.e. (7/10)*(6/9) + (7/10) * (3/9) = 21/30 = 7/10
But the answer is D 14/15.
This is got by 1 - probability of picking both the marbles as red
i.e. (3/10) * (2/9) = 1/15
Therefore 1 - 1/15 = 14/15
Can anybody explain why the answers differ. Thanks.
