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A jar holds 12 balls, each of which is colored white, black, gray, or
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26 Aug 2019, 00:00
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
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26 Aug 2019, 00:27
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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Re: A jar holds 12 balls, each of which is colored white, black, gray, or
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26 Aug 2019, 00:32
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)=1P(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
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26 Aug 2019, 01:22
White + Black + Gray + Green = 12 and Let
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A jar holds 12 balls, each of which is colored white, black, gray, or
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Updated on: 27 Aug 2019, 01:22
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
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26 Aug 2019, 08:15
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 whiteW, blackB, greyY, & greenG 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. +1 kudos if you find my post helpful.
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Re: A jar holds 12 balls, each of which is colored white, black, gray, or
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26 Aug 2019, 08:53
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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Re: A jar holds 12 balls, each of which is colored white, black, gray, or
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26 Aug 2019, 08:58
(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
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26 Aug 2019, 09:05
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
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26 Aug 2019, 09:26
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
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26 Aug 2019, 18:30
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 Answer is E



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Re: A jar holds 12 balls, each of which is colored white, black, gray, or
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26 Aug 2019, 20:15
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
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