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Math Expert V
Joined: 02 Sep 2009
Posts: 58401
A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 75% (01:37) correct 25% (01:46) wrong based on 59 sessions

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Competition Mode Question

A jar holds 12 balls, each of which is colored white, black, gray, or green. If a person is to select a ball randomly from the jar, is the probability less than 1/2 that the ball selected will be either white or green?

(1) The probability that the person will select a gray ball is 1/3.

(2) The probability that the person will select a green ball is 1/6.

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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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1
Given: A jar holds 12 balls, each of which is colored white(W), black(B), gray(Gy), or green(Gn).
W+B+Gy+Gn = 12 (1)

Asked: If a person is to select a ball randomly from the jar, is the probability less than 1/2 that the ball selected will be either white or green?
Q. Is W+Gn<6

(1) The probability that the person will select a gray ball is 1/3.
Gy = 4
W+B+Gn = 8
W+Gn < 8
NOT SUFFICIENT

(2) The probability that the person will select a green ball is 1/6.
Gn = 2
W+B+Gy=10
W+Gn <12
NOT SUFFICIENT

Combining (1) & (2)
(1) The probability that the person will select a gray ball is 1/3.
Gy = 4
(2) The probability that the person will select a green ball is 1/6.
Gn = 2
W + B = 6
W <6
W + Gn < 8
NOT SUFFICIENT

IMO E
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Senior Manager  G
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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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1
We know that 12 balls in a jar are colored either White(w), black(b), green(gn), or gray(gy).
We are to determine if P(w) + P(gn) < 1/2.

P(w)+P(gn)+P(gy)+P(b)=1
So P(w)+P(gn)=1-P(gy)-P(b)
We therefore need either P(w) and P(gn), or P(gy) and P(b).

From 1, know only P(gy)=1/3. This is insufficient.
From 2, we know only P(gn)=1/6. This is also insufficient, since clearly P(gn) is less than 1/2. It’s possible adding P(b) to 1/6 could be less than 1/2 or more than 1/2.

1+2 still not sufficient because we need either P(gy) and P(b), or P(w) and P(gn). What we have is P(gy) and P(gn). Hence insufficient.

Answer is E imo.

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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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1
White + Black + Gray + Green = 12 and Let
Attachments A jar holds 12 balls.JPG [ 199.89 KiB | Viewed 630 times ]

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A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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A jar holds 12 balls, each of which is colored white, black, gray, or green. If a person is to select a ball randomly from the jar, is the probability less than 1/2 that the ball selected will be either white or green?

(1) The probability that the person will select a gray ball is 1/3.

(2) The probability that the person will select a green ball is 1/6.

total = 12 balls
If a person is to select a ball randomly from the jar, is the probability less than 1/2 that the ball selected will be either white or green

#1The probability that the person will select a gray ball is 1/3.

gray must be 4 balls ; insufficient

#2The probability that the person will select a green ball is 1/6.
green must be 2 balls
insufficient
from 1 & 2
total balls of white & black ; 6
so P of white and green can be <6 ; not possible to determine
IMO E

Originally posted by Archit3110 on 26 Aug 2019, 06:24.
Last edited by Archit3110 on 27 Aug 2019, 01:22, edited 1 time in total.
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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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A jar holds 12 balls, each of which is colored white, black, grey, or green. If a person is to select a ball randomly from the jar, is the probability less than 1/2 that the ball selected will be either white or green?

Let white-W, black-B, grey-Y, & green-G

To find: $$\frac{(W + G)}{12}$$ < $$\frac{1}{2}$$

(1) The probability that the person will select a grey ball is 1/3.
No information about other cards. INSUFFICIENT!

(2) The probability that the person will select a green ball is 1/6.
No information about white cards. INSUFFICIENT!

(1) + (2)
Y= $$\frac{1}{3}$$ * 12 =4
and,
G= $$\frac{1}{6}$$ * 12 =2

Remaining cards = 6 (W & B), but we still don't know show how much each W & B cards are there in the jar. It can be any number (<6) so the probability will keep changing. Hence, INSUFFICIENT!

IMO answer is option E.

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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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IMO E
Is white/green <1/2

White 3/6, Green 1/6, Blue 0, Grey 2/6 ---- No
White 0, Green 1/6, Blue 3/6, Grey 2/6---- Yes
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GMAT 1: 650 Q46 V33 GMAT 2: 660 Q45 V35 Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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(1) The probability that the person will select a gray ball is 1/3.
Hence the number of Gray balls = (1/3)12 = 4.
No info is given about the number of green and white balls, therefore insufficient.

(2) The probability that the person will select a green ball is 1/6.
Number of Green balls = (1/6)12 = 2.
There is no info about the number/probability of white balls, therefore insufficient.

If we combine both the statements, we still would have no info about the number of white balls.

Hence the answer would be E
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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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Pmax = 8/12
Pmin= 1/6

Hence E is the answer

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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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A jar holds 12 balls, each of which is colored white, black, gray, or green. If a person is to select a ball randomly from the jar, is the probability less than 1/2 that the ball selected will be either white or green?

(1) The probability that the person will select a gray ball is 1/3.

(2) The probability that the person will select a green ball is 1/6.

The question is asking that, whether there are less than 6 W+G balls?

Info 1: THere are 4 gray balls. Clearly this is not sufficient to conclude the number of W+Green balls. they can be as many as 7(prob more than 1/2) or as few as 2(prob less than 1/2).

Info 2: There are 2 green balls. Again this is not enough to identify the probability as white can be as many as 8 and as few as 1.

Together. WE know that there are 4 gray balls. 2 Green balls. Still this is not enough to conclude the probability of W+Green balls.
As white can be as many as 5 and as few as 1. If white is 5, the probability (7/12>1/2) and if white is 1, the probability is(1/4<1/2).

Hence E.
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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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A jar holds 12 balls, each of which is colored white, black, gray, or green. If a person is to select a ball randomly from the jar, is the probability less than 1/2 that the ball selected will be either white or green?
White balls = w
Black balls = b
Gray balls = gy
Green balls= gn
Given that w+b+gr+gr = 12
We need to find out how many white and green balls are there in the jar ?

(1) The probability that the person will select a gray ball is 1/3.
P(gr) = 1/3, so from this we can infer that there are 4 (probability of gray balls= 4/12 = 1/3) gray balls in the jar, however we can't find how many white and green balls are there in the jar, not sufficient

(2) The probability that the person will select a green ball is 1/6.
P(gn) = 1/6, from this we can infer that there are 2 green in the jar, again, we can't find out how many white and green balls are there in jar, not sufficient

(1)&(2) gy = 4, gn = 2, w+b = 6
Still, we can't find how many white and green balls are there in the jar, not sufficient
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Re: A jar holds 12 balls, each of which is colored white, black, gray, or  [#permalink]

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I ) points out probability of gray s 1/3 hence their are total of 4 gray balls
Thus sum of other two balls = 8
thus probability of selecting one of them is 8/12 which si greater than 1/2
Sufficient
II) green ball is 1/6.
hence green balls = 2 now white can be less than or greater than 4
thus no sufficient
Hence A is the answer Re: A jar holds 12 balls, each of which is colored white, black, gray, or   [#permalink] 26 Aug 2019, 20:15
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