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Probability

by [email protected] » Mon Sep 12, 2011 9:19 am
Q: THERE ARE 4 BALLS IN A BAG OF COLOR RED, BLUE, YELLOW AND GREEN ? IF TWO BALLS ARE TO BE SELECTED
WHAT IS THE PROBABILITY THAT AT LEAST ONE BLUE OR ONE GREEN BALL WILL BE SELECTED ?

1: 1/2 2: 1/16 3: 2/6 3: 5/6 4: 3/6
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by shankar.ashwin » Mon Sep 12, 2011 9:39 am
I am getting 7/8, but dont see it in the answer choices. I am assuming there is only 1 ball of each color.

So, to find prob. of getting either a blue or green,

I try to find the probability of getting the other 2 colored balls and subtracting the value with one

i.e; Prob of getting a Red or Yellow is 2/4, followed by getting the other color is 1/4.

So Probability of getting a Red followed by Yellow or the other way around is 1/2*1/4= 1/8,

So prob of getting either a blue or green should be 1- 1/8 = 7/8.

Maybe I am missing something here.

[email protected] wrote:Q: THERE ARE 4 BALLS IN A BAG OF COLOR RED, BLUE, YELLOW AND GREEN ? IF TWO BALLS ARE TO BE SELECTED
WHAT IS THE PROBABILITY THAT AT LEAST ONE BLUE OR ONE GREEN BALL WILL BE SELECTED ?

1: 1/2 2: 1/16 3: 2/6 3: 5/6 4: 3/6
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by cans » Mon Sep 12, 2011 9:41 am
r,b,y,g.
total ways=4C2 = 6.
at least 1 b or 1 g means 6 - no b and no g. = 6 - r and y = 6-1 = 5.
Prob = 5/6
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by shankar.ashwin » Mon Sep 12, 2011 9:45 am
An easy way to understand this problem would be to consider the case where we do not get both blue or green balls in the 2 we pick up. In other words, the 2 balls we pick are yellow and red. This is one unique case.

All other cases would include either the blue or green. Total outcomes are 4C2=6.
Case where we have no blue and green = 1.
All other cases would include either the blue or green - 5 such cases. So 5/6
shankar.ashwin wrote:I am getting 7/8, but dont see it in the answer choices. I am assuming there is only 1 ball of each color.

So, to find prob. of getting either a blue or green,

I try to find the probability of getting the other 2 colored balls and subtracting the value with one

i.e; Prob of getting a Red or Yellow is 2/4, followed by getting the other color is 1/4.

So Probability of getting a Red followed by Yellow or the other way around is 1/2*1/4= 1/8,

So prob of getting either a blue or green should be 1- 1/8 = 7/8.

Maybe I am missing something here.

[email protected] wrote:Q: THERE ARE 4 BALLS IN A BAG OF COLOR RED, BLUE, YELLOW AND GREEN ? IF TWO BALLS ARE TO BE SELECTED
WHAT IS THE PROBABILITY THAT AT LEAST ONE BLUE OR ONE GREEN BALL WILL BE SELECTED ?

1: 1/2 2: 1/16 3: 2/6 3: 5/6 4: 3/6
Last edited by shankar.ashwin on Tue Sep 13, 2011 1:49 am, edited 1 time in total.
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by Deepthi Subbu » Tue Sep 13, 2011 1:10 am
cans wrote:r,b,y,g.
total ways=4C2 = 6.
at least 1 b or 1 g means 6 - no b and no g. = 6 - r and y = 6-1 = 5.
Prob = 5/6
IMO C
Hi cans ,

How did you find r to be 1 ?
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