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Please explain:

Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.

Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.

1.) 2/5 2.) 6/10 3.) 7/10 4.) 15/20 5.) 19/20

Vote for 3)

Andy pick 2 out of 5 without replacing the marbles back into the bag. The chances of picking up 2 blue marbles are 3/5*1/2=3/10, thus the probability pikcing up at least 1 red marble is 1-3/10=7/10.

can you see why we have different results, while our approaches are directly reverse? You come from behind and I'm from the front. Both correct. Can't see the problem.

Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.

1.) 2/5 2.) 6/10 3.) 7/10 4.) 15/20 5.) 19/20

P(Atleast one red) = 1- probablity (no red)
= 1 - 3/5
= 2/5

P(Atleast one red) = 1- probablity (no red) = 1 - 3/5 = 2/5

Krisrini, it's a smart way of doing it..

I calculated this as Probability that both are red + Probability that one is read = 2/5*1/4 + 2/5 * 3/4 = 2/20 + 6/20 = 2/5

It should be 7/10.

Probability of no reds is not 3/5. It is 3/5 *2/4 = 3/10
1 - Prob of red = 1 - 3/10 = 7/10

Different approach:
- There are 5 balls, thus there are 5C2 ways to pick 2, or (5*4)/2! = 10.
- There are 3C2 ways to pick only blues (3*2)/2! = 3.
- (Total combinations) - (combinations with no reds) = 10 - 3 = 7
- Therefore, 7/10 is the prob of picking at least 1 red.