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A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two

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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

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New post 30 Nov 2014, 05:33
Ralphcuisak wrote:
In a similar question.

In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball?

a) 1/5.
b) ½.
c) 1/3.
d) 3/10.
e) ¼.


the OA is 1/2..which is just 1/10=1/5*x,so x=1/2..
my question is that returning the ball does not figure into the question in anyway?


The probability of drawing a red ball is 1/5: P = (red)/(total) = 1/5;
The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10: (red)/(total)*(white)/(total) = 1/10.

Since (red)/(total) = 1/5, then 1/5*(white)/(total) = 1/10 --> (white)/(total) = 1/2.

Answer: B.

Hope it's clear.
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

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New post 12 Mar 2015, 09:04
nevergiveup wrote:
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

1- [3!/(2!1!)]/[10!/(2!8!)]=14/15

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

[3!/(2!1!)]/[10!/(2!8!)]=1/15


3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15


Personally, I really don't know what is the difference between question 3 & 4.

Having trouble with this. For the first question the way that I am attempting to reason through it is 10!/2!8! . 2! represents the 2 marbles selected and 8! is for the other 8 NOT selected. I then have probability first one is not blue ,3/10, times this and answer subtracted from 1. So 1- [10!/2!/8!(3/10)] :x
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

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New post 12 Mar 2015, 09:12
GMAT01 wrote:
nevergiveup wrote:
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

1- [3!/(2!1!)]/[10!/(2!8!)]=14/15

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

[3!/(2!1!)]/[10!/(2!8!)]=1/15


3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15


Personally, I really don't know what is the difference between question 3 & 4.

Having trouble with this. For the first question the way that I am attempting to reason through it is 10!/2!8! . 2! represents the 2 marbles selected and 8! is for the other 8 NOT selected. I then have probability first one is not blue ,3/10, times this and answer subtracted from 1. So 1- [10!/2!/8!(3/10)] :x


hi GMAT01,
you do not require the coloured portion above...
wherever the questions of this form are there , the best is to take probability none is there and it seems you begun that way but went wrong mid way..
prob of both not being blue means both are red..
prob of first red=3/10 and second red=2/9.....
prob of both red 3/10*2/9=6/90..
so prob of atleast one blue=1-6/90=84/90...
hope its clear
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

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New post 14 Nov 2016, 10:50
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
For above question, can you show me how to solve by
1- p(all blue)
as I am getting different answer by doing 1- (7/10*6/9) = 8/15
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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

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Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two   [#permalink] 04 Oct 2018, 11:43

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