A jar holds 12 balls, each of which is colored white, black, gray, or

White + Black + Gray + Green = 12 and Let

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

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.

+1 kudos if you find my post helpful.

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

(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

Pmax = 8/12
Pmin= 1/6

Hence E is the answer

Posted from my mobile device

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.

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

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