Probability

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Probability

by akpareek » Mon Apr 26, 2010 7:56 am
There are 8 marbles in a jar. 5 are red and 3 are green. If 4 marbles are selected at random from jar, what is the probability that at least one red marble and one green marble will be seleted?

A) 4/5 B) 10/11 C) 11/12 D) 12/13 E) 13/14
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by iamseer » Mon Apr 26, 2010 11:55 am
The ways in which at least one red marble and one green marble can be selected are
1R,3G or 2R,2G or 3R,1G (these are the only possibilities)

so IMO answer is 2/3

Let me know what you think.
Thanks.
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by clock60 » Mon Apr 26, 2010 12:23 pm
akpareek wrote:There are 8 marbles in a jar. 5 are red and 3 are green. If 4 marbles are selected at random from jar, what is the probability that at least one red marble and one green marble will be seleted?

A) 4/5 B) 10/11 C) 11/12 D) 12/13 E) 13/14
i got E, my reasoning
the total number of ways to select 4 marbles out of 8 =8C4=5*7*2=70
now we have several options
3-red,1-green
2-red, 2-green
1-red, 3 green
i did not find otners
5C3*3C1+5C2*3C2+5C1*3C3=30+30+5=65
the desired result=65/70=13/14

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by iamseer » Tue Apr 27, 2010 12:11 am
clock60 wrote: i got E, my reasoning
the total number of ways to select 4 marbles out of 8 =8C4=5*7*2=70
now we have several options
3-red,1-green
2-red, 2-green
1-red, 3 green
i did not find otners
5C3*3C1+5C2*3C2+5C1*3C3=30+30+5=65
the desired result=65/70=13/14
Yes, you are absolutely correct. Same reasoning.
I must have been out of mind to arrive at the number 2/3. :)

Correct answer is indeed 13/14
"Choose to chance the rapids and dance the tides"