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Probability "with replacement"

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by mmukher » Sat Mar 22, 2008 5:36 pm
5. An urn contains 7 red and 4 blue balls. Find the corresponding probabilities if the balls are replaced after each draw.

(a) both red
(b) a red and a blue
(c) both the same colour.


I can solve for "without replacement" but any ideas for with replacement?
For eg
5a. I think should be 7/11 * 7/11 = 49/121 ...

Is that the right approach?

Thanks !!
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by madhavi » Sun Mar 23, 2008 6:25 am
Yes, mmukher. Your approach is right.

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by simplyjat » Sun Mar 23, 2008 6:43 am
Replacement implies how many balls you have for second and consecutive draws...
The approach you have mentioned is one with replacement... for without replacement, 7/11 * 6/10...
simplyjat
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by freedsl » Wed Mar 26, 2008 5:06 pm
(a) both red
(b) a red and a blue
(c) both the same colour.
so for a and b above, since they are independent events, we say P ( Red & Red) for example is P(R)*P(R).

But how do we do c, is this correct.

P(c) = P (Both RED or Both Blue) = P(Both Blue) + P(Both Red) -P(both Blue)* P(Both Red)

Is this correct??
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by richardwang6430 » Tue Apr 15, 2008 5:59 pm
Yes, you are right
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by GMAT_crusher » Tue Apr 15, 2008 6:26 pm
richardwang6430 wrote:Yes, you are right
I did not quite get the P(c)..

if we need both Red OR both Blue
then
P(c) = P(both Red) + P (bothBlue)
= 49/121 + 16/121
= 65/121

What did I miss?
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by anshul265 » Wed Apr 16, 2008 5:50 pm
I think this is the correct method. When we need AND we multiply and when we need OR we add.

So to solve the question:

P(c)= 7/11*6/10 + 4/11*3/10
= 42/110 + 12/110
= 54/110
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by nysnowboard » Tue Mar 09, 2010 12:05 am
Hey all. First of all, sorry to revive a dead thread but I wanted to make sure my thought process was satisfactory when it comes to probability, my major weakness.

This is my first post and hopefully I don't step on any toes....

restating the question:

5. An urn contains 7 red and 4 blue balls. Find the corresponding probabilities if the balls are replaced after each draw.

(a) both red
(b) a red and a blue
(c) both the same colour.


a.) Since the balls are replaced after each draw, this is simply 7/11 * 7/11 = 49/121

b.) This should be 7/11 * 4/11 = 28/121

c.)As for P(c), I'm a little hazy. If we get both the same color, shouldn't this be a mutually exclusive case? (You can't get an outcome of both red AND both blue since there are only two draws).

Mutual exclusion (P(two red and two blue) = 0, so P(two red or two blue) = P(two red) + P(two blue) = 49/121 + 16/121 = 65/121

Could someone who really knows the correct strategy correct me? =)
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by sanju09 » Tue Mar 09, 2010 1:06 am
nysnowboard wrote:Hey all. First of all, sorry to revive a dead thread but I wanted to make sure my thought process was satisfactory when it comes to probability, my major weakness.

This is my first post and hopefully I don't step on any toes....

restating the question:

5. An urn contains 7 red and 4 blue balls. Find the corresponding probabilities if the balls are replaced after each draw.

(a) both red
(b) a red and a blue
(c) both the same colour.


a.) Since the balls are replaced after each draw, this is simply 7/11 * 7/11 = 49/121

right

b.) This should be 7/11 * 4/11 = 28/121

right

c.)As for P(c), I'm a little hazy. If we get both the same color, shouldn't this be a mutually exclusive case? (You can't get an outcome of both red AND both blue since there are only two draws).

Mutual exclusion (P(two red and two blue) = 0, so P(two red or two blue) = P(two red) + P(two blue) = 49/121 + 16/121 = 65/121
Could someone who really knows the correct strategy correct me? =)

right

and you are most welcome
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by nysnowboard » Tue Mar 09, 2010 11:23 pm
Thanks for the input, it's truly valued =)
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by Mrgharios » Sat Sep 24, 2016 5:23 am
Part B is false. You should multiply the result by 2 because you can get either 1 red and then 1 blue or 1 blue and then 1 red
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