A box contains 11 balls, of which 6 are yellow and 5 are blue. If 2 ba

Question

A box contains 11 balls, of which 6 are yellow and 5 are blue. If 2 balls are chosen at random without replacement, what is the probability that 1 will be yellow and 1 will be blue?

A: 30/121
B: 3/11
C: 1/2
D: 6/11
E: 3/5

LucienH · Accepted Answer

Probability of picking first a yellow then a blue: 
6/11 * 5/10 (since no replacement)

Probability of picking first a blue and then a yellow: 
5/11 * 6/10

Combined probability = 30/110 + 30/110 = 6/11

ANS D

Archit3110 · Answer

without replacement
6/11 * 5/10 * 2c1
6/11 
option D

chetan2u · Answer

Three ways
1) Find the opposite and subtract from total.
So P that both balls are of same colour =>  \frac{6C2}{11C2}+\frac{5C2}{11C2}=\frac{6*5+5*4}{11*10}=\frac{50}{110} 
Thus our answer =  1-\frac{5}{11}=\frac{6}{11}

2) P that first yellow and then blue=  \frac{6C1*5C1}{11P2}=\frac{3}{11} 
P that first blue and then yellow=  \frac{5C1*6C1}{11P2}=\frac{3}{11} 
Total =  \frac{6}{11}

3) P that first yellow and then blue=  \frac{6}{11}*\frac{5}{10}=\frac{3}{11} 
P that first blue and then yellow= \frac{5}{11}*\frac{6}{10}=\frac{3}{11} 
Total =  \frac{6}{11}

D

Score-Quest · Answer

We can consider the two possible scenarios in which one yellow and one blue ball can be chosen:

Choosing a yellow ball first and then blue.
Choosing a blue ball first and then yellow.

Let's calculate the probability for each scenario:

Probability of choosing a yellow ball first and then a blue ball:
Probability of choosing a yellow ball first = 6/11 (since there are 6 yellow balls out of 11 total balls)
After choosing a yellow ball, there are now 10 balls left, of which 5 are blue.
Probability of choosing a blue ball next = 5/10 = 1/2
So, the probability of this scenario is: (6/11) * (1/2) = 6/22 = 3/11

Probability of choosing a blue ball first and then a yellow ball:
Probability of choosing a blue ball first = 5/11 (since there are 5 blue balls out of 11 total balls)
After choosing a blue ball, there are now 10 balls left, of which 6 are yellow.
Probability of choosing a yellow ball next = 6/10 = 3/5
So, the probability of this scenario is: (5/11) * (3/5) = 15/55 = 3/11

Total probability = 3/11 + 3/11 = 6/11

sayan640 · Answer

The highlighted portion is wrong.. 6.5 / 11 c 2 -> 6/11   chetan2u   Bunuel   bb

chetan2u · Answer

Thank you. 11C2 was supposed to be a 11P2

brivibas39tv · Answer

I had this question in Mock 4 and answered 3/11. I seem to always forget to multiply by the amount of ways to chose the balls, this time blue-yellow and yellow-blue (2C1).

Does anybody know a good resource where I can study this so I do not make the same mistake again?

MartyMurray · Answer

A box contains 11 balls, of which 6 are yellow and 5 are blue. If 2 balls are chosen at random without replacement, what is the probability that 1 will be yellow and 1 will be blue?

Probability = Favorable/Total

Total = 11c2 = 55

Favorable = 6c1 × 5c1 = 6 × 5 = 30

30/55 = 6/11

A: 30/121
B: 3/11
C: 1/2
D: 6/11
E: 3/5

Correct answer:   D

GMATinsight · Answer

Favorable outcomes = 6C1 * 5C1 = 30

Total Outcomes of choosing two balls out of 11 = 11C2 = 55

Required Probability = 30/55 = 6/11

Answer: Option D

Related Video:  
  https://www.youtube.com/watch?v=G7a_MBqZjTY

A box contains 11 balls, of which 6 are yellow and 5 are blue. If 2 ba

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 11 balls, of which 6 are yellow and 5 are blue. If 2 ba

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