Probability "with replacement"

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mmukher: Senior | Next Rank: 100 Posts; Posts: 34; Joined: Wed Feb 20, 2008 7:15 pm; Thanked: 6 times; Followed by:1 members

Probability "with replacement"

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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Source: Beat The GMAT — Problem Solving | Sat Mar 22, 2008 5:36 pm

madhavi: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Tue Jun 26, 2007 11:33 pm; Thanked: 4 times

by madhavi » Sun Mar 23, 2008 6:25 am

Yes, mmukher. Your approach is right.

https://goal-mba.blocked

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simplyjat: Master | Next Rank: 500 Posts; Posts: 423; Joined: Thu Dec 27, 2007 1:29 am; Location: Hyderabad, India; Thanked: 36 times; Followed by:2 members; GMAT Score:770

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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freedsl: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Wed Mar 26, 2008 10:30 am; Location: so cal

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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richardwang6430: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Sat Apr 12, 2008 6:23 pm; Thanked: 1 times

by richardwang6430 » Tue Apr 15, 2008 5:59 pm

Yes, you are right

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GMAT_crusher: Junior | Next Rank: 30 Posts; Posts: 25; Joined: Mon Mar 31, 2008 6:09 pm

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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anshul265: Senior | Next Rank: 100 Posts; Posts: 71; Joined: Thu Jan 31, 2008 7:01 am; Thanked: 2 times

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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nysnowboard: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Sat Mar 06, 2010 6:07 am; Thanked: 1 times

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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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

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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