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# In a jar there are 3 red balls and 2 blue balls. What is the

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Manager
Joined: 24 Mar 2010
Posts: 75
In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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31 Jul 2014, 06:47
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70% (01:20) correct 30% (01:41) wrong based on 165 sessions

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In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

A. 9/10
B. 16/20
C. 2/5
D. 3/5
E. ½

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Math Expert
Joined: 02 Sep 2009
Posts: 59730
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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31 Jul 2014, 07:00
1
LucyDang wrote:
In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

A. 9/10
B. 16/20
C. 2/5
D. 3/5
E. ½

P(at least one red) = 1 - P(no red, so 2 blue) = 1- 2/5*1/4 = 9/10.

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Math Expert
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Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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31 Jul 2014, 07:01
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Manager
Joined: 24 Mar 2010
Posts: 75
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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31 Jul 2014, 07:33
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Bunuel wrote:

Hi Bunuel,

You're super fast!!! Many thanks to your resources! Indeed, I'm practicing probability right now, hihi..
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Manager
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In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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31 Jul 2014, 20:40
Another way to solve this problem is:

P (at least 1 red) = 1 - P (no red)

$$P(H=2)=\frac{C^2_2}{C^2_5}=\frac{1}{10}$$

Hence, $$P({at least 1 red})=1-\frac{1}{10}=\frac{9}{10}$$.
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Manager
Joined: 11 Jun 2014
Posts: 52
Concentration: Technology, Marketing
GMAT 1: 770 Q50 V45
WE: Information Technology (Consulting)
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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31 Jul 2014, 21:15
1
P ( at least one red) = 1 - P ( all blue balls )

P( all blue balls ) = 2/5 ( for the first draw) * 1/4 ( for the second draw)
= 2/20 = 1/10

P ( at least one red ) = 1 - 1/10 = 9/10
Intern
Joined: 03 Oct 2011
Posts: 14
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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05 Aug 2014, 18:57
I see that using $$P(at least 1 red)$$ = $$1 - P(no red)$$ is the quick way to solve this.

However, should $$P(at least 1 red) = P(1 red) + P(2 red)$$ work?

where $$P(2 red) = \frac{3C2}{5C2}$$ and $$P(1 red)= \frac{2(3C1*2C1)}{5C2}$$

Its not working ofcource because $$P(1 red)= \frac{2(3C1*2C1)}{5C2}$$ is greater than 1. I multiplied by 2 because I think I have seen Bunuel solve such problems and account for the two ways of picking two balls i.e (B,R) or (R,B)

Where am I wrong in my thinking?

Thanks for the help
Intern
Joined: 14 Dec 2013
Posts: 3
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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05 Aug 2014, 19:33
gooner wrote:
I see that using $$P(at least 1 red)$$ = $$1 - P(no red)$$ is the quick way to solve this.

However, should $$P(at least 1 red) = P(1 red) + P(2 red)$$ work?

where $$P(2 red) = \frac{3C2}{5C2}$$ and $$P(1 red)= \frac{2(3C1*2C1)}{5C2}$$

Its not working ofcource because $$P(1 red)= \frac{2(3C1*2C1)}{5C2}$$ is greater than 1. I multiplied by 2 because I think I have seen Bunuel solve such problems and account for the two ways of picking two balls i.e (B,R) or (R,B)

Where am I wrong in my thinking?

Thanks for the help

Hi Gooner,

Here order of choosing Blue and Red do not matter. Hence do not multiply with 2.
$$P(1 red)= \frac{(3C1*2C1)}{5C2}$$
$$P(2 red)= \frac{(3C2)}{5C2}$$

Finally probability of choosing atleast one red is P(1 red) + p(2 red) = $$\frac{(3C1*2C1)}{5C2}$$ + $$\frac{(3C2)}{5C2}$$.

i.e., 3+6/10 = 9/10.
Intern
Joined: 03 Oct 2011
Posts: 14
In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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05 Aug 2014, 19:55
Sowmith wrote:
gooner wrote:
I see that using $$P(at least 1 red)$$ = $$1 - P(no red)$$ is the quick way to solve this.

However, should $$P(at least 1 red) = P(1 red) + P(2 red)$$ work?

where $$P(2 red) = \frac{3C2}{5C2}$$ and $$P(1 red)= \frac{2(3C1*2C1)}{5C2}$$

Its not working ofcource because $$P(1 red)= \frac{2(3C1*2C1)}{5C2}$$ is greater than 1. I multiplied by 2 because I think I have seen Bunuel solve such problems and account for the two ways of picking two balls i.e (B,R) or (R,B)

Where am I wrong in my thinking?

Thanks for the help

Hi Gooner,

Here order of choosing Blue and Red do not matter. Hence do not multiply with 2.
$$P(1 red)= \frac{(3C1*2C1)}{5C2}$$
$$P(2 red)= \frac{(3C2)}{5C2}$$

Finally probability of choosing atleast one red is P(1 red) + p(2 red) = $$\frac{(3C1*2C1)}{5C2}$$ + $$\frac{(3C2)}{5C2}$$.

i.e., 3+6/10 = 9/10.

Thanks for that. I am trying to think of the scenario where I saw the solution account for the two ways of selecting the desired outcome.
Intern
Joined: 28 Dec 2015
Posts: 37
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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13 Jul 2016, 03:55
3 Red and 2 Blue balls

Balls taken out can be RR or RB

RR=Number of combinations=2!/2!
RB=Number of combinations=2!(RB or BR)

P(RB)+P(RR)=2*3/5*2/4+3/5*2/4=9/10
Director
Joined: 20 Feb 2015
Posts: 737
Concentration: Strategy, General Management
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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13 Jul 2016, 08:19
p(no red balls drawn)=2/5*1/4=1/10
p(at least 1 red)=1-1/10=9/10
Manager
Status: IF YOU CAN DREAM IT, YOU CAN DO IT
Joined: 03 Jul 2017
Posts: 186
Location: India
Concentration: Finance, International Business

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09 Jul 2017, 17:47
In a jar there are 3 red balls and 2 blue balls. What is the probability og drawing at least one red ball when drawing two consecutive balls randomly.
A.9/10
B.16/20
C.2/5
D.3/5
E.1/2

Why are we considering this question to be without replacement ?? Can someone please explain under what conditions do we have to do so? Thank you
Director
Joined: 05 Mar 2015
Posts: 978

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09 Jul 2017, 19:00
longhaul123 wrote:
In a jar there are 3 red balls and 2 blue balls. What is the probability og drawing at least one red ball when drawing two consecutive balls randomly.
A.9/10
B.16/20
C.2/5
D.3/5
E.1/2

Why are we considering this question to be without replacement ?? Can someone please explain under what conditions do we have to do so? Thank you

Because he draws two consecutive balls randomly. therefore with replacement could not be considered...
Senior Manager
Joined: 28 Jun 2015
Posts: 279
Concentration: Finance
GPA: 3.5

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09 Jul 2017, 19:08
Usually if the ball is replaced it will either be mentioned specifically, or you will be given that the ball was drawn "one by one" or "one after another", instead of consecutively.
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Posts: 53
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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24 Nov 2019, 00:57
Bunuel wrote:
LucyDang wrote:
In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

A. 9/10
B. 16/20
C. 2/5
D. 3/5
E. ½

P(at least one red) = 1 - P(no red, so 2 blue) = 1- 2/5*1/4 = 9/10.

Bunuel

I understood that the best way to solve these "Atleast" problem is to find ( 1 - P) way.
But i proceeded by actually finding the Probabilities as given below:
Probability of drawing both red balls = 3/5 x 2/4 =6/20 = 3/10
Probability of drawing one red & one blue ball= 3/5 x 2/4 = 3/10 ( even if we draw blue ball first, the final value remains the same. = 2/5 x 3/4 =6/20 = 3/10)

Now adding both the probabilities, 3/10 + 3/10 = 2 x 3/10 = 6 /10 = 3/5 which is answer D

Can you tell where i went wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 59730
Re: In a jar there are 3 red balls and 2 blue balls. What is the  [#permalink]

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24 Nov 2019, 02:04
pandajee wrote:
Bunuel wrote:
LucyDang wrote:
In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

A. 9/10
B. 16/20
C. 2/5
D. 3/5
E. ½

P(at least one red) = 1 - P(no red, so 2 blue) = 1- 2/5*1/4 = 9/10.

Bunuel

I understood that the best way to solve these "Atleast" problem is to find ( 1 - P) way.
But i proceeded by actually finding the Probabilities as given below:
Probability of drawing both red balls = 3/5 x 2/4 =6/20 = 3/10
Probability of drawing one red & one blue ball= 3/5 x 2/4 = 3/10 ( even if we draw blue ball first, the final value remains the same. = 2/5 x 3/4 =6/20 = 3/10)

Now adding both the probabilities, 3/10 + 3/10 = 2 x 3/10 = 6 /10 = 3/5 which is answer D

Can you tell where i went wrong.

P(at least one red) = P(1 red, 1 blue) + P(2 red) = 3/5*2/4*2 + 3/5*2/4 = 9/10. You should multiply 3/5*2/4 by 2 because 1 red and 1 blue can occur in two ways: RB or BR.
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Re: In a jar there are 3 red balls and 2 blue balls. What is the   [#permalink] 24 Nov 2019, 02:04
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