A box contains 20 balls, of which 12 are red and 8 are blue. If two

Question

A box contains 20 balls, of which 12 are red and 8 are blue. If two balls are to be drawn from this box at random without replacement, what is the probability that one ball will be red and the other will be blue?

A 1/96
B 6/25
C 24/95
D 48/95
E 1

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imeanup · Answer

Number of ways to draw  1   red ball is   {^{12}}C_1

Number of ways to draw  1  blue ball is  ^{8}C_1

Therefore, total number of ways to draw one red ball and one blue ball  = {^{12}}C_1 	imes {^{8}}C_1

extit{Probability} = \dfrac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}}

Probability =  \dfrac{{^{12}}C_1 	imes {^{8}}C_1}{{^{20}}C_2} = \dfrac{12 	imes 8}{10 	imes 19} = \dfrac{48}{95}

D

Akkiiii · Answer

Number of ways to draw 1 red ball is =12 C 1

Number of ways to draw 1 blue ball is= 8 C 1

Therefore, total number of ways to draw one red ball and one blue ball  without replacement = 20C1 * 19C1= 20*19

Probability=Number of favorable outcomes/Total number of outcomes

Probability = 12*8/20*19= 24/95

C

poojaarora1818 · Answer

A box contains 20 balls, of which 12 are red and 8 are blue. If two balls are to be drawn from this box at random without replacement, what is the probability that one ball will be red and the other will be blue?

A 1/96
B 6/25
C 24/95
D 48/95
E 1

Solution

Probability Formula = Favorable outcomes/Total outcomes
 
We know the total no. of balls in a box = 20
No. of red balls in a box                      = 12
No. of blue balls in a box                     = 8

So,   12C1 * 8C1/20C2                        = 48/95

Option D

imeanup · Answer

Another way to approach this problem when combinatorics is not known is simply by considering the probability of each choice.

In the first draw, we have:
 
 First blue, then red (or) (or),  P_{B 	o R} = \dfrac{8}{20} 	imes \dfrac{12}{19}  
 First red, then blue,  P_{R 	o B} = \dfrac{12}{20} 	imes \dfrac{8}{19}

Total Probability  P = \dfrac{12	imes 8 + 8 	imes 12}{20 	imes 19} = \dfrac{48}{95}

HadjiMurad · Answer

why not 24/95? There is not question about which one is first blue or red, just 2 different balls, what's probability that one of them blue and another one is red. ???

Bunuel · Answer

That's a common mistake when solving with the probability approach. The key point is that the scenario of picking one red and one blue ball (the BR scenario) can occur in two different orders: red first, then blue (RB), or blue first, then red (BR):

(8/20 * 12/19) + (12/20 * 8/19).

A box contains 20 balls, of which 12 are red and 8 are blue. If two

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GMAT Problem Solving (PS) Questions

A box contains 20 balls, of which 12 are red and 8 are blue. If two

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