Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 26 May 2017, 15:28

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Each of 25 balls in a box are either blue, red, white and

Author Message
Manager
Joined: 19 Sep 2005
Posts: 110
Followers: 1

Kudos [?]: 11 [0], given: 0

Each of 25 balls in a box are either blue, red, white and [#permalink]

### Show Tags

10 Oct 2005, 09:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

1. The probability that it will be both white and even is 0
2. the probability that it will be white minus the probability that it will be even is .2

Manager
Joined: 03 Oct 2005
Posts: 87
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

10 Oct 2005, 09:24
Yikes! Is the answer to this question possible?

Is this a Data Sufficiency question?

(I think I've forgotten all my probability background. Will have to go searching for some site that covers it. Anybody have a nice site?)
VP
Joined: 22 Aug 2005
Posts: 1113
Location: CA
Followers: 1

Kudos [?]: 111 [0], given: 0

### Show Tags

10 Oct 2005, 09:26
I will chose E.

n = 25

p(W or Even) = P(W) + P(Even) - P (W and Even)

(1) P(W and Even) = 0
as we dont know indivisual probabilities. Insufficient.

(2) P(W) - p(Even) = 0.2
Insufficient.

together, insufficient as we still cannot find P(w) + p(even) to get P(W or Even)
Manager
Joined: 19 Sep 2005
Posts: 110
Followers: 1

Kudos [?]: 11 [0], given: 0

### Show Tags

10 Oct 2005, 09:40

if the prob. of getting white and even is 0, this means that either the prob. of getting white OR the prob. of getting even is 0 since it's one * the other.
right?

Going with this, move on to B. If white minues even is .2 and we know one of them has a probability of 0, then why can't we say that the probabiliity of either is .2?
Manager
Joined: 06 Oct 2005
Posts: 88
Location: Beantown
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

10 Oct 2005, 10:00
I got (E)

I. Here you are only dealing with prob white * prob even white. This says there are either no white balls or no even white balls. We're looking for either white OR even of any color. All the other balls might be even, or all but 1 etc.

II. The prob white - Prob even could work for many instances. With out knowing anything about the amounts of the other balls.

Combining them does not help at all.

B.
Manager
Joined: 19 Sep 2005
Posts: 110
Followers: 1

Kudos [?]: 11 [0], given: 0

### Show Tags

10 Oct 2005, 10:12
Still not convinced why it can't be C!
Manager
Joined: 06 Oct 2005
Posts: 88
Location: Beantown
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

10 Oct 2005, 10:34
Because you are falling for the trap in I. it is simply stating there are no white + even balls. It has nothing to do with the overall probability of choosing an even numbered ball.

There is a global probability of choosing even or odd just as there is a global probability of choosing each color. But within each color group there is a further probability of choosing even or odd.

We are concerned with the global probability of choosing even and I. just gives us the % of evens in the white group, what if all the blues were even and 2/3 the reds even too?

make sense?

B.
VP
Joined: 22 Aug 2005
Posts: 1113
Location: CA
Followers: 1

Kudos [?]: 111 [0], given: 0

### Show Tags

10 Oct 2005, 11:20
Jennif102 wrote:

if the prob. of getting white and even is 0, this means that either the prob. of getting white OR the prob. of getting even is 0 since it's one * the other.
right?

P(A and B) = P(A) * P(B)

This is ONLY true iff both A & B are Independent. Otherwise, it would be:
P(A and B) = P (A) * P(B/A) = P(B) * P(A/B)

from the question stem we do not know both A and B are independent.
Manager
Joined: 19 Sep 2005
Posts: 110
Followers: 1

Kudos [?]: 11 [0], given: 0

### Show Tags

10 Oct 2005, 11:54
great explanation guys! I understand now. Thanks!
got it!   [#permalink] 10 Oct 2005, 11:54
Display posts from previous: Sort by