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Please correct me if i am wrong.

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 a) exactly one is blue? b) at least one marble is blue?

desired / total a) 3C1 * 7C1 / 10C2 = 21/45

b) Total - P(none) / total 10C2 - 3C2 / 10C2 = 45-3/45 = 42/45 = 14/15
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

isnt the probability that at least one is blue equal to the sum of the probability that you have one blue, one red and the probability that both are blue ?

isnt the probability that at least one is blue equal to the sum of the probability that you have one blue, one red and the probability that both are blue ?

Yes, but that's more work than finding finding out the probability of 2 red and just subtracting that from 1.

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 a) exactly one is blue? b) at least one marble is blue?

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 a) exactly one is blue? b) at least one marble is blue?

desired / total a) 3C1 * 7C1 / 10C2 = 21/45

Getting 7/15 for a.

p(exactly 1 blue) = 7/10 * (1-6/9) * 2! = 7/15

thats the same value. 21/45 = 7/15
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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 a) exactly one is blue? b) at least one marble is blue?

desired / total a) 3C1 * 7C1 / 10C2 = 21/45

b) Total - P(none) / total 10C2 - 3C2 / 10C2 = 45-3/45 = 42/45 = 14/15

A: consider two possiblities: 3/10*7/9 (2) --> 21/45 --> 7/15 b/c we have to consider the scenario of 7/10*3/9. So the multiplied by 2 is the same as 21/90+21/90 --> 42/90 --> 7/15.

B: 1-prob. that none will be blue. 3/10*2/9 --> 6/90 --> 15/15-1/15 --> 14/15

Walker is right for qn1. for exactly one to be blue, it should be BR & RB 7/10*3/8 + 3/10*7/9 = 7/15 for qn2. atleast one Blue, you can add BB to the above & get the answer. prob of BB is 7/10*6/9 =7/15 so total of BB< RB and BR = 2*7/15= 14/15