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A box contains 10 red balls and 30 blue balls. Balls will be pulled ou
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07 Oct 2021, 09:29
07 Oct 2021, 09:29
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Question Stats:
81% (02:02) correct
18% (02:04) wrong based on 11 sessions
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A box contains 10 red balls and 30 blue balls. Balls will be pulled out of the box and then put back. It will stop when 2 blue balls have been pulled out of the box. What is the probability of getting at least 1 red ball before stopping?
A. 1/2
B. 2/3
C. 9/16
D. 3/4
E. 7/16
A. 1/2
B. 2/3
C. 9/16
D. 3/4
E. 7/16
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Re: A box contains 10 red balls and 30 blue balls. Balls will be pulled ou
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07 Oct 2021, 10:45
07 Oct 2021, 10:45
1
Carcass wrote:
A box contains 10 red balls and 30 blue balls. Balls will be pulled out of the box and then put back. It will stop when 2 blue balls have been pulled out of the box. What is the probability of getting at least 1 red ball before stopping?
A. 1/2
B. 2/3
C. 9/16
D. 3/4
E. 7/16
A. 1/2
B. 2/3
C. 9/16
D. 3/4
E. 7/16
We want P(at least 1 red ball before stopping)
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(at least 1 red ball before stopping) = 1 - P(NO red balls before stopping)
In other words, P(at least 1 red ball before stopping) = 1 - P(blue balls only)
P(blue balls only)
P(blue balls only) = P(1st ball is blue AND 2nd ball is blue)
= P(1st ball is blue) x P(2nd ball is blue)
= 30/40 x 30/40
= 3/4 x 3/4
= 9/16
So, P(at least 1 red ball before stopping) = 1 - 9/16
= 7/16
Answer: E
Cheers,
Brent
General Discussion
A box contains 10 red balls and 30 blue balls. Balls will be pulled ou
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11 Oct 2021, 03:49
11 Oct 2021, 03:49
GreenlightTestPrep wrote:
Carcass wrote:
A box contains 10 red balls and 30 blue balls. Balls will be pulled out of the box and then put back. It will stop when 2 blue balls have been pulled out of the box. What is the probability of getting at least 1 red ball before stopping?
A. 1/2
B. 2/3
C. 9/16
D. 3/4
E. 7/16
A. 1/2
B. 2/3
C. 9/16
D. 3/4
E. 7/16
We want P(at least 1 red ball before stopping)
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(at least 1 red ball before stopping) = 1 - P(NO red balls before stopping)
In other words, P(at least 1 red ball before stopping) = 1 - P(blue balls only)
P(blue balls only)
P(blue balls only) = P(1st ball is blue AND 2nd ball is blue)
= P(1st ball is blue) x P(2nd ball is blue)
= 30/40 x 30/40
= 3/4 x 3/4
= 9/16
So, P(at least 1 red ball before stopping) = 1 - 9/16
= 7/16
Answer: E
Cheers,
Brent
So we assume the ball is replaced?
