Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?
8/28
9/28
10/28
10/18
11/18
PS, tough probability
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- jayhawk2001
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- hemanth28
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Different possible combinations of colors is
G R B
3 2 0
3 0 2
3 1 1
2 1 2
2 2 1
2 0 3
1 2 2
1 1 3
0 2 3
these are the possible colour combinations when 5 balls are picked.
out of this the color combinaton required is 2 1 2
hence the probability is 1/9.
Can someone tell me if i am missing something?
G R B
3 2 0
3 0 2
3 1 1
2 1 2
2 2 1
2 0 3
1 2 2
1 1 3
0 2 3
these are the possible colour combinations when 5 balls are picked.
out of this the color combinaton required is 2 1 2
hence the probability is 1/9.
Can someone tell me if i am missing something?
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hemanth28 wrote:Different possible combinations of colors is
G R B
3 2 0
3 0 2
3 1 1
2 1 2
2 2 1
2 0 3
1 2 2
1 1 3
0 2 3
these are the possible colour combinations when 5 balls are picked.
out of this the color combinaton required is 2 1 2
hence the probability is 1/9.
Can someone tell me if i am missing something?
dude ur missing out the various ways the req. trio can occur..always follow combinatorics fr such type of questions..
answer is 9/28
Abhishek sunku
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Total number of ways you can pick 5 balls out of 8 is 8C5 (Order is not important).
Out of those combinations the favourable events are where you have 1 red, 2g and 3 blue.
And you can pick up 1r, 2g and 2b in 2C1 * 3C2 * 3C2 ways given (you have 2 red, 3 green and 3 blue).
I think that's how he got the equation.
Hope this helps.
Calista.
Out of those combinations the favourable events are where you have 1 red, 2g and 3 blue.
And you can pick up 1r, 2g and 2b in 2C1 * 3C2 * 3C2 ways given (you have 2 red, 3 green and 3 blue).
I think that's how he got the equation.
Hope this helps.
Calista.
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That's not the only way to pick up the balls now, is it?
For e.g. you are assuming you get a blue ball first, then another blue, then a green, another green and then a red (i.e. b,b,g,g,r).
But what about the other possibilities? like g,g,b,b,r or b,g,r,b,g etc?
If you add all of those then you MUST get the same probability.
Obviously, this isn't a faster approach.
Calista.
For e.g. you are assuming you get a blue ball first, then another blue, then a green, another green and then a red (i.e. b,b,g,g,r).
But what about the other possibilities? like g,g,b,b,r or b,g,r,b,g etc?
If you add all of those then you MUST get the same probability.
Obviously, this isn't a faster approach.
Calista.
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ukr.net already revealed the OA as B earlier in this thread...ukr.net wrote:the answer is B 9/28
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The statement says "...He randomly picks 5 from the bag without replacement" and I understand how the answer to this question is calculated.
But I am wondering what would be the answer if the statement said instead "WITH REPLACEMENT after piking each ball"??? :
But I am wondering what would be the answer if the statement said instead "WITH REPLACEMENT after piking each ball"??? :
The statement says "...He randomly picks 5 from the bag without replacement" and I understand how the answer to this question is calculated.
But I am wondering what would be the answer if the statement said instead "WITH REPLACEMENT after piking each ball"??? :
But I am wondering what would be the answer if the statement said instead "WITH REPLACEMENT after piking each ball"??? :