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There are 3 black balls and 7 white balls in a box. If two balls are c
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24 May 2022, 01:48
24 May 2022, 01:48
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There are 3 black balls and 7 white balls in a box. If two balls are chosen randomly, what is the probability that at least one will be black?
(A) 1/15
(B) 7/30
(C) 7/15
(D) 8/15
(E) 23/30
(A) 1/15
(B) 7/30
(C) 7/15
(D) 8/15
(E) 23/30
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022)
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GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
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Re: There are 3 black balls and 7 white balls in a box. If two balls are c
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24 May 2022, 07:07
24 May 2022, 07:07
Carcass wrote:
There are 3 black balls and 7 white balls in a box. If two balls are chosen randomly, what is the probability that at least one will be black?
(A) 1/15
(B) 7/30
(C) 7/15
(D) 8/15
(E) 23/30
(A) 1/15
(B) 7/30
(C) 7/15
(D) 8/15
(E) 23/30
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 ball is black) = 1 - P(not getting at least 1 black ball)
What does it mean to not get at least 1 black ball? It means getting zero black balls.
So, we can write: P(getting at least 1 black ball) = 1 - P(getting zero black balls)
P(getting zero black balls)
P(getting zero black balls) = P(1st ball is white AND 2nd ball is white)
= P(1st ball is white) x P(2nd ball is white)
= 7/10 x 6/9
= 42/90
= 7/15
So, P(getting at least 1 black ball) = 1 - 7/15
=15/15 - 7/15
= 8/15
Answer: D
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There are 3 black balls and 7 white balls in a box. If two balls are c
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24 May 2022, 11:41
24 May 2022, 11:41
1
Remember that for any "at least" Probability question, you'll want to do this:
total - what we don't want
In other words, that's...
1 - no blacks
The next step is to calculate the situation where we get neither to be black; remember that this is dependent because the selection of the first ball will affect the number of balls in the box:
P(W)*P(W)
(7/10) * (6/9) = 7/10 * 2/3 = 7/5 * 1/3 = 7/15
Don't forget to do the subtraction at this point!
Total - no blacks
1 - P(W)*P(W)
1 - 7/15 = 8/15
The answer is D.
Best,
Rowan
PS There are loads of Probability resources in my signature! Check them out...
total - what we don't want
In other words, that's...
1 - no blacks
The next step is to calculate the situation where we get neither to be black; remember that this is dependent because the selection of the first ball will affect the number of balls in the box:
P(W)*P(W)
(7/10) * (6/9) = 7/10 * 2/3 = 7/5 * 1/3 = 7/15
Don't forget to do the subtraction at this point!
Total - no blacks
1 - P(W)*P(W)
1 - 7/15 = 8/15
The answer is D.
Best,
Rowan
PS There are loads of Probability resources in my signature! Check them out...
