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

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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, 06:47
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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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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
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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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31 Jul 2014, 07:01
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31 Jul 2014, 07:33
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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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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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
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
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Posts: 16
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
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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.
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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.
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Joined: 28 Dec 2015
Posts: 39
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
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Posts: 795
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
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Joined: 03 Jul 2017
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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: 999

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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: 290
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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Re: probability   [#permalink] 09 Jul 2017, 19:08
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