In a box there are A green balls, 3A + 6 red balls and 2 yellow ones.

Question

In a box there are A green balls, 3A + 6 red balls and 2 yellow ones. If there are no other colors, what is the probability of taking out a green or a yellow ball?

A 1/5.
B 1/2.
C 1/3.
D 1/4.
E 2/3.

BrentGMATPrepNow · Accepted Answer

Since none of the answer choices contain the variable A, we can safely  assign a nice value to the value of A

If A = 1, then we have: 
1 green ball
9 red balls
2 yellow balls
So, there are  12  balls altogether, and  3  of the balls are either green or yellow

So, P(selected balls is either green or yellow) =  3 / 12  
= 1/4

Answer: D

Cheers.
Brent

BrentGMATPrepNow · Answer

We can also use algebra.

GIVEN: 
A = number of green balls
3A + 6 = number of red balls
2 = number of yellow balls
TOTAL number of balls = A + (3A + 6) + 2 =  4A + 8 
Number of balls that are either green or yellow =  A + 2

So, P(selected balls is either green or yellow) =  (A + 2) / (4A + 8)  
=  (1)(A + 2) / (4)(A + 2)  
= 1/4

Answer: D

Cheers.
Brent

Harsh9676 · Answer

I think the information anout number of green balls is missing in the question.

Without that we cannot get the total no of balls and cannot find the probability.

The question just said there are green balls, 3A+2 red and 2 yellow.

Total = green balls + 3A+2+ 2

Posted from my mobile device

philipssonicare · Answer

Probability space is G + R + Y and we want the probability of G+Y
 \frac{(G+Y)}{(G+R+Y)}

G=A
R=3A+6
Y=2

\frac{(A+2)}{(4A+8)}

\frac{(A+2)}{ } 
 \frac{1}{ }=\frac{1}{4}

shridhar786 · Answer

green balls = A

red balls = 3A+6

yellow balls = 2

Total numbers of ball = 4A+8

we have to find the probability of taking out a green or a yellow ball

since both, the events are mutually exclusive (ones that cannot happen at the same time)

probability 0f taking out (a green or a yellow) = probability of taking out (a green) + probability of taking out (a yellow)

probability 0f taking out (a green or a yellow) =  \frac{green ball}{total number of ball}  +  \frac{yellow ball}{total number of ball}

probability 0f taking out (a green or a yellow) =  \frac{A}{4A+8} + \frac{2}{4A+8}

probability 0f taking out (a green or a yellow) =  \frac{(A+2)}{4(A+2)}  =  \frac{1}{4}

D is the answer

bumpbot · Answer

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In a box there are A green balls, 3A + 6 red balls and 2 yellow ones.

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