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 !!
Probability "with replacement"
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- simplyjat
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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...
The approach you have mentioned is one with replacement... for without replacement, 7/11 * 6/10...
simplyjat
so for a and b above, since they are independent events, we say P ( Red & Red) for example is P(R)*P(R).(a) both red
(b) a red and a blue
(c) both the same colour.
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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I did not quite get the P(c)..richardwang6430 wrote:Yes, you are right
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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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? =)
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? =)
- sanju09
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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
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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