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Lily has only red and blue balls in a jar. If she removes three balls,

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Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

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New post 10 Jun 2018, 07:21
1
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

35% (02:11) correct 65% (02:43) wrong based on 91 sessions

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Lily has only red and blue balls in a jar. If she removes three balls, is the probability of getting all red balls greater than the probability of getting at least one blue ball?

(1) The number of red balls is more than three times the number of blue balls.
(2) Less than (1)/(4) th of the balls in the jar are blue.

Source: Expert Global

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Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

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New post 10 Jun 2018, 08:50
Answer should be E. Eg: Take there are 4 red balls and 1 blue ball. Then P(RRR)= 2/5 and P(atleast 1B) = 3/5. So 2nd expression is smaller.

Then take there are 7 red balls and 1 blue ball. Then P(RRR)= 5/8 and P(atleast 1B) = 3/8. So 1st expression is smaller.

This example will work for both the statements. Therefore, no sufficiency in any. Hence, E.
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Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

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New post 28 Aug 2018, 10:13
Hi. How did you get 3/5 for blue balls?
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Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

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New post 29 Aug 2018, 05:37
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Hi
I find the resource helpful, you can also try this probability lecture.
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Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

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New post 29 Aug 2018, 05:48
Thanks so much for this

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Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

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New post 29 Aug 2018, 06:57
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Quote:
Lily has only red and blue balls in a jar. If she gets three balls randomly, is the probability of getting all red balls greater than the probability of getting at least one blue ball?

(1) The number of red balls is more than three times the number of blue balls.
(2) Less than (1)/(4) of the balls in the jar are blue.

Source: Expert Global


Nice problem!

\(\operatorname{P} \left( {{\text{all}}\,\,3\,\,{\text{red}}} \right)\,\,\,\,\mathop > \limits^? \,\,\,\operatorname{P} \left( {{\text{not}}\,\,{\text{all}}\,\,3\,\,{\text{red}}} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\operatorname{P} \left( {{\text{all}}\,\,3\,\,{\text{red}}} \right) = \boxed{\frac{{C\left( {r,3} \right)}}{{C\left( {r + b,3} \right)}}\,\,\,\,\mathop > \limits^? \,\,\,\frac{1}{2}}\)

\(\left( {1 + 2} \right)\,\,\)

\(Take\,\,\left( {r,b} \right) = \left( {4,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {4,3} \right)}}{{C\left( {5,3} \right)}} = \,\frac{1}{{10}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle\)

\(Take\,\,\left( {r,b} \right) = \left( {5,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {5,3} \right)}}{{C\left( {6,3} \right)}} = \,\frac{{10}}{{20}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,{\text{but}}\,\,\,{\text{increasing}}\,\,...\)

\(Take\,\,\left( {r,b} \right) = \left( {6,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\frac{{C\left( {6,3} \right)}}{{C\left( {7,3} \right)}} = \,\frac{4}{7}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\,\)

(Keeping the number of blue balls fixed and increasing the number of red ones, we know the probability of getting 3 reds must increase...)

The answer is (E), because we were able to BIFURCATE (1+2)!

The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Re: Lily has only red and blue balls in a jar. If she removes three balls,  [#permalink]

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Re: Lily has only red and blue balls in a jar. If she removes three balls,   [#permalink] 23 Nov 2019, 20:55
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