A box contains exactly 24 balls, of which 12 are red and 12 are blue.

Question

A box contains exactly 24 balls, of which 12 are red and 12 are blue. If two balls are to be picked from this box at random and without replacement, what is the probability that both balls will be red?

(A) 11/46
(B) 1/4
(C) 5/12
(D) 17/40
(E) 19/40

(A) 11/46
(B) 1/4
(C) 5/12
(D) 17/40
(E) 19/40

generis · Accepted Answer

Straightforward: 
 \frac{favorable.outcome}{possible.outcomes}  for 2 picks, then multiply.

Pick #1: 12 red balls. 24 balls total 
 Probability of red in 1st pick :  \frac{12}{24}=\frac{1}{2}

Now there are 11 red and 23 total balls. (There's no replacement.)

Probability of red in 2nd pick :  \frac{11}{23}

Multiply probability of both picks.
P(2 successive reds):
 (\frac{1}{2}*\frac{11}{23}) =\frac{11}{46}

Answer A

GMATinsight · Answer

12C2 / 24C2 = 12*11/24*23 = 11/46

Answer: Option A

nipunjain14 · Answer

To choose 2 red balls among red balls : 12C2 = 6*11
To choose 2 balls out of 24 balls : 24C2 = 12*23

total probablity of choosing 2 red balls without replcement = 6*11/12*23 = 11/46
Ans :A

pushpitkc · Answer

Given data : 24 balls in the box - 12 red and 12 blue

To find : Probability that both balls are red

Denominator(2 balls chosen from 24 balls)  24c2 = \frac{24*23}{2} = 12*23 
Numerator(2 red balls from 12 available)  12c2 = \frac{12*11}{2} = 6*11

Probability that 2 red balls are chosen(without replacement) =  \frac{6*11}{12*23} = \frac{11}{46}  (Option A)

sonikavadhera · Answer

Simple probability:

probability first ball red: 12/24
probability second ball red: 11/23

total probability : 1/2*11/23 = 11/46 
Option A

+1 kudos if u like the post.

TimeTraveller · Answer

\frac{12c2}{24c2} = \frac{12*11}{24*23} = \frac{11}{46} . Ans - A.

zilchzn · Answer

could someone give an explanation to why is this approach wrong and correct my thinking

approach 1. Selecting 2 red balls - 12c1 * 11c1 
total - 24c1 * 23c1

instead of

approach 2. 12c2/24c2

although it comes up to the same thing however if i mix up 1 and 2 say 12c1*11c1/24c2 - it doesn't work.

I incorrectly took this approach and came up with 11/23 which is wrong but would like to know why we shouldn't mix these 2 approaches.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

A box contains exactly 24 balls, of which 12 are red and 12 are blue.

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A box contains exactly 24 balls, of which 12 are red and 12 are blue.

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards