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

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Bunuel
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A box contains 20 balls, of which 12 are red and 8 are blue. If two [#permalink] New post   29 Mar 2024, 01:30
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­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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meanup
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A box contains 20 balls, of which 12 are red and 8 are blue. If two [#permalink] New post   29 Mar 2024, 02:28
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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 \(= {^{12}}C_1 \times {^{8}}C_1\)

\(\textit{Probability} = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)

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

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A box contains 20 balls, of which 12 are red and 8 are blue. If two [#permalink] New post   30 Mar 2024, 00:25
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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), \(P_{B \to R} = \dfrac{8}{20} \times \dfrac{12}{19}\)
  • First red, then blue, \(P_{R \to B} = \dfrac{12}{20} \times \dfrac{8}{19}\)­

Total Probability \(P = \dfrac{12\times 8 + 8 \times 12}{20 \times 19} = \dfrac{48}{95}\)­
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Re: A box contains 20 balls, of which 12 are red and 8 are blue. If two [#permalink] New post   29 Mar 2024, 13:09
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­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

 
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Re: A box contains 20 balls, of which 12 are red and 8 are blue. If two [#permalink] New post   29 Mar 2024, 11:16
Bunuel wrote:
­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



This is a PS Butler Question

­

­

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
 
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Re: A box contains 20 balls, of which 12 are red and 8 are blue. If two [#permalink]
29 Mar 2024, 11:16
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