Each of the 25 balls in a certain box is either red, blue or

[amk250]

Trying to understand why the answer is E.


Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

1) The probability that the ball will both be white and have an even number painted on it is 0.
2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2.

[StaceyKoprince]

25 balls
each one is red, white, or blue
each one has a number from 1 to 10

Want: white OR even (note that we DON'T want white AND even - we have to be able to strip out those that fall into both categories). To calculate, can either do:
a) probability of "white and odd" + probability of "even and not white"
b) probability of white + probability of even - probability of white & even

(1) Translated, this means there aren't any that are both white and even. This doesn't tell us how many are white or how many are even. Insufficient. Eliminate A and D.

(2) Pwhite - Peven = 0.2. So, Pwhite could be 0.4 which would make Peven 0.2. Or Pwhite could be 0.3 which would make Peven 0.1. And (by itself) it doesn't tell me Prob of even & white, which I'd need to subtract, so... insufficient in many ways. Eliminate B.

(1) AND (2) Now I know that Peven+white = 0. BUT, I still have multiple possibilities for Pwhite and Peven (see above). 0.4+0.2-0=0.6. 0.3+0.1-0=0.4. ?? Still insufficient.

[arielle.bertman]

Hi Stacey

Can you explain how you derived option a) P(white and odd) + P(even and not white)?

[ashish.jere]

Thanks Stacy.

[Ben Ku]

Glad it helped!

[i.ahmed111]

Hi Stacey

I am also confused on how you derived option a) P(white and odd) + P(even and not white)? Wouldn't we be missing a section that we should be including? (white AND even?)

thanks!

[RonPurewal]

here's an easier way.

since there are 25 balls in the box, you don't need to use probability at all. you can just translate all the probabilities into numbers of balls. to do so, just multiply them by 25.

also, note that we don't care about red vs. blue, since the only thing that's mentioned is white vs. not white. so, we can use a DOUBLE SET MATRIX..

--

here's the translated question:

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. How many balls are either white or even-numbered?

1) There are no balls that are both white and even-numbered.

MATRIX:
------------- EVEN / NOT EVEN / TOTAL
WHITE ------- 0 ----- blank ------ blank
NOT WHITE - blank -- blank ----- blank
TOTAL ------ blank -- blank ------- 25
insufficient

2) The number of white balls minus the number of even-numbered balls is 5.

so if X even numbered balls, then (X + 8) white balls.
MATRIX:
------------- EVEN / NOT EVEN / TOTAL
WHITE ------- blank -- blank ---- X + 5
NOT WHITE - blank -- blank ----- 20 - X
TOTAL -------- X ----- 25 - X ------- 25
(note that the italics have been calculated by subtraction)
insufficient

--

TOGETHER
MATRIX:
------------- EVEN / NOT EVEN / TOTAL
WHITE -------- 0 ----- X + 5 ---- X + 5
NOT WHITE -- X ----- 20 - 2X --- 20 - X
TOTAL -------- X ----- 25 - X ------- 25
(note that the italics have been calculated by subtraction. EITHER THE ROW OR THE COLUMN GIVES THE SAME EXPRESSION)
X could be anywhere from 0 to 10. (if x is more than 10, then 20 - 2x becomes negative.)
still insufficient.

so (e)
[editor: i fixed an arithmetic mistake. everything appearing in boldface has been corrected. thanks to the poster who pointed out the mistake.]

[sharmin.karim]

in Quote 2 how did you get 8 as the difference between number of white balls and number of even-numbered balls?
[editor: stupid mistake. i fixed it; see above. thanks for catching it.]

[esledge]

Hi Sharmin,

I get 5. The 8 above was probably caused by a simple computation error.

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2.

Prob of white - prob of even = 0.2 = 5/25
number of white - number of even = 5

[sharmin.karim]

Thanks Emily. I had gotten 5 as well.

[RonPurewal]

Hi Sharmin,

I get 5. The 8 above was probably caused by a simple computation error.

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2.

Prob of white - prob of even = 0.2 = 5/25
number of white - number of even = 5

thanks. i corrected everything in that post.

[mycshops]

25 balls
each one is red, white, or blue
each one has a number from 1 to 10

Want: white OR even (note that we DON'T want white AND even - we have to be able to strip out those that fall into both categories). To calculate, can either do:
a) probability of "white and odd" + probability of "even and not white"
b) probability of white + probability of even - probability of white & even

When the GMAT says "has a number from 1 to 10 painted on it" could this mean any frequency of any of the numbers from 1 to 10? or can we assume this means each number from 1 to 10 will be used just once as the balls are sequentially numbered balls?

[RonPurewal]

25 balls
each one is red, white, or blue
each one has a number from 1 to 10

Want: white OR even (note that we DON'T want white AND even - we have to be able to strip out those that fall into both categories). To calculate, can either do:
a) probability of "white and odd" + probability of "even and not white"
b) probability of white + probability of even - probability of white & even

When the GMAT says "has a number from 1 to 10 painted on it" could this mean any frequency of any of the numbers from 1 to 10? or can we assume this means each number from 1 to 10 will be used just once as the balls are sequentially numbered balls?

it means that any one of those numbers could be painted on any one of the balls.
by the way, you can tell that this has to be the case, since there are 25 balls -- it wouldn't be possible to paint only the numbers 1-10 consecutively on 25 balls!

[tsiria]

I just solved this problem using properties of sets:
P(A) or P(B) = P(A) + P(B) - P(both A&B)

We are told in S1: P(A&B) = 0, we still need P(A) and p(B), so S1 is insufficient

S2: gives us: P(A) - P(B), but we need P(A)+(B), so S2 is also insufficient. from s2, A and B can take multiple values.

Combining both S1 and S2 does not still provide p(A)+p(B), hence
answer is E.

Note: A stands for white, B is Even number

[tsiria]

I just solved this problem using properties of sets:
P(A) or P(B) = P(A) + P(B) - P(both A&B)

We are told in S1: P(A&B) = 0, we still need P(A) and p(B), so S1 is insufficient

S2: gives us: P(A) - P(B), but we need P(A)+(B), so S2 is also insufficient. from s2, A and B can take multiple values.

Combining both S1 and S2 does not still provide p(A)+p(B), hence
answer is E.

Note: A stands for white, B is Even number

I figured I could use the properties of sets based on the type of information given in the statements.