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A box contains 10 red balls and 30 blue balls. Balls will be pulled

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Intern
Joined: 25 Oct 2016
Posts: 4
A box contains 10 red balls and 30 blue balls. Balls will be pulled  [#permalink]

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Updated on: 07 Oct 2018, 05:40
2
00:00

Difficulty:

35% (medium)

Question Stats:

67% (01:42) correct 33% (02:12) wrong based on 53 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

Originally posted by trankimphuong on 07 Oct 2018, 02:11.
Last edited by chetan2u on 07 Oct 2018, 05:40, edited 1 time in total.
Corrected the choices
Math Expert
Joined: 02 Aug 2009
Posts: 8326
A box contains 10 red balls and 30 blue balls. Balls will be pulled  [#permalink]

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07 Oct 2018, 05:39
1
trankimphuong 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. 9/6
D. 3/4
E. 7/6

Choices are clearly flawed. You can't have probability >1. Both C(9/6) and E(7/6) are >1..
So it should be 16 in denominator..

Easiest way is to find prob of both being BLUE, meaning NO red .
Probability of B one after another = $$\frac{30}{10+30}*\frac{30}{10+30}=\frac{3*3}{4*4}=\frac{9}{16}$$
This is the only way that Red is not picked up before the game stops. Otherwise the ways can be
1) B,R,R,B
2)R,R,B,B
3) R,R,R,R,R,B,R,B
And so on

So probability that atleast one red is picked = 1-(probability that NO red is picked)=1-9/16=7/16

E
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Joined: 22 Feb 2018
Posts: 413
Re: A box contains 10 red balls and 30 blue balls. Balls will be pulled  [#permalink]

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07 Oct 2018, 06:51
1
trankimphuong 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

OA:E

Probability of getting at least 1 red ball before stopping $$= 1- \frac{30}{40}*\frac{30}{40}=1-\frac{9}{16}=\frac{7}{16}$$
Re: A box contains 10 red balls and 30 blue balls. Balls will be pulled   [#permalink] 07 Oct 2018, 06:51
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