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parveen110
- Senior | Next Rank: 100 Posts
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Veritas prep question.
There are three red balls and two black balls in a jar. What is the probability
that John gets at least one red ball when he picks two balls at random out of
the jar at the same time?
The answer is: 9/10
My approach:
I am considering complementary events of taking no red balls.
I may pick up two balls in 5C2 ways.
and then, first black ball may be picked up in 2C1 ways and the other one in 1C1 ways.
Therefore, (2C1*1C1)/(5C2)=1/5
so, probablity of picking one red ball is: 1-(1/5)=4/5.
Where's my approach flawed?
There are three red balls and two black balls in a jar. What is the probability
that John gets at least one red ball when he picks two balls at random out of
the jar at the same time?
The answer is: 9/10
My approach:
I am considering complementary events of taking no red balls.
I may pick up two balls in 5C2 ways.
and then, first black ball may be picked up in 2C1 ways and the other one in 1C1 ways.
Therefore, (2C1*1C1)/(5C2)=1/5
so, probablity of picking one red ball is: 1-(1/5)=4/5.
Where's my approach flawed?
