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# Three squares on a chessboard are chosen at random. The probability

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Joined: 23 Sep 2015
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Three squares on a chessboard are chosen at random. The probability  [#permalink]

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06 Apr 2019, 23:37
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Three squares on a chessboard are chosen at random. The probability that 2 of them are of one color and the remaining one is of another color is:

A) 3/64
B) 64/441
C) 5/21
D) 8/21
E) 16/21

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Joined: 02 Aug 2009
Posts: 8320
Three squares on a chessboard are chosen at random. The probability  [#permalink]

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07 Apr 2019, 07:55
2
aragonn wrote:
Three squares on a chessboard are chosen at random. The probability that 2 of them are of one color and the remaining one is of another color is:

A) 3/64
B) 64/441
C) 5/21
D) 8/21
E) 16/21

When looking for probability, don't get worked up on probability and combinations..
Just be sure that you choose same for both favorable and total.

(I) combinations, that is order does not matter..
Favorable outcomes - 32C2 for first color and 32C1 for second, and both colors can be used twice, so 32C2*32C1*2=$$\frac{32*31*32*2}{2}$$
Total outcomes - 64C3 =$$\frac{64*63*62}{3*2}$$
Thus, Probability = $$\frac{32C2*32C1*2}{64C2}=\frac{32*31*32*3}{64*63*62}=\frac{16}{21}$$

(II) Permutations, that is order matters..
Favorable outcomes - 32P2 for first color and 32P1 for second, and both colors can be chosen in $$\frac{3!}{2}=3$$ ways, so 32*31*32*3
Total outcomes - 64P3 =$$64*63*62$$
Thus, Probability =$$\frac{32*31*32*3}{64*63*62}=\frac{16}{21}$$

E
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Re: Three squares on a chessboard are chosen at random. The probability  [#permalink]

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08 Oct 2019, 06:52
chetan2u wrote:
aragonn wrote:
Three squares on a chessboard are chosen at random. The probability that 2 of them are of one color and the remaining one is of another color is:

A) 3/64
B) 64/441
C) 5/21
D) 8/21
E) 16/21

When looking for probability, don't get worked up on probability and combinations..
Just be sure that you choose same for both favorable and total.

(I) combinations, that is order does not matter..
Favorable outcomes - 32C2 for first color and 32C1 for second, and both colors can be used twice, so 32C2*32C1*2=$$\frac{32*31*32*2}{2}$$
Total outcomes - 64C3 =$$\frac{64*63*62}{3*2}$$
Thus, Probability = $$\frac{32C2*32C1*2}{64C2}=\frac{32*31*32*3}{64*63*62}=\frac{16}{21}$$

(II) Permutations, that is order matters..
Favorable outcomes - 32P2 for first color and 32P1 for second, and both colors can be chosen in $$\frac{3!}{2}=3$$ ways, so 32*31*32*3
Total outcomes - 64P3 =$$64*63*62$$
Thus, Probability =\frac{32*31*32*3}{64*63*62}=\frac{16}{21}[/m]

E

Permutations - Why 2 colors can be chosen in 3!/2? Please explain. Thank you.
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Joined: 16 Jan 2019
Posts: 507
Location: India
Concentration: General Management
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Re: Three squares on a chessboard are chosen at random. The probability  [#permalink]

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08 Oct 2019, 07:21
1
Note that

Probability that 2 of them are of one color and the remaining one is of another color = 1 - (Probability that all three are of the same color)

Probability that all are white = $$(\frac{32}{64})(\frac{31}{63})(\frac{30}{62})=\frac{5}{42}$$

Similarly, probability that all are black = $$\frac{5}{42}$$

Therefore, probability that all three chosen squares are of the same color= $$\frac{5}{42}+\frac{5}{42}=\frac{5}{21}$$

And so, probability that 2 of them are of one color and the remaining one is of another color $$=1-(\frac{5}{21})=\frac{16}{21}$$

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Joined: 02 Aug 2009
Posts: 8320
Re: Three squares on a chessboard are chosen at random. The probability  [#permalink]

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08 Oct 2019, 08:37
htmtruong wrote:
chetan2u wrote:
aragonn wrote:
Three squares on a chessboard are chosen at random. The probability that 2 of them are of one color and the remaining one is of another color is:

A) 3/64
B) 64/441
C) 5/21
D) 8/21
E) 16/21

When looking for probability, don't get worked up on probability and combinations..
Just be sure that you choose same for both favorable and total.

(I) combinations, that is order does not matter..
Favorable outcomes - 32C2 for first color and 32C1 for second, and both colors can be used twice, so 32C2*32C1*2=$$\frac{32*31*32*2}{2}$$
Total outcomes - 64C3 =$$\frac{64*63*62}{3*2}$$
Thus, Probability = $$\frac{32C2*32C1*2}{64C2}=\frac{32*31*32*3}{64*63*62}=\frac{16}{21}$$

(II) Permutations, that is order matters..
Favorable outcomes - 32P2 for first color and 32P1 for second, and both colors can be chosen in $$\frac{3!}{2}=3$$ ways, so 32*31*32*3
Total outcomes - 64P3 =$$64*63*62$$
Thus, Probability =\frac{32*31*32*3}{64*63*62}=\frac{16}{21}[/m]

E

Permutations - Why 2 colors can be chosen in 3!/2? Please explain. Thank you.

Hi, there are 3 squares and two of them have same colour, that is why we do 3!/2..
Say there are 3 square A, B and C and there are two colours Red and White to be filled, so TWo will have one colour and third will have the second colour..

The way to choose combination of 3 squares when two are same is 3!/2, where division by 2 is for the duplicity of colour.
If all three were different colours, it would be 3!
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Three squares on a chessboard are chosen at random. The probability  [#permalink]

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08 Oct 2019, 10:01
aragonn wrote:
Three squares on a chessboard are chosen at random. The probability that 2 of them are of one color and the remaining one is of another color is:

A) 3/64
B) 64/441
C) 5/21
D) 8/21
E) 16/21

Given: Three squares on a chessboard are chosen at random.
Asked: The probability that 2 of them are of one color and the remaining one is of another color is:

Total ways of picking 3 squares on a chessboard = 64C3 = 64*63*62/6

Number of ways of picking 2 of them are of one color and the remaining one is of another color = 32C2*32C1*2 = 32*31*32

Probability = 6*32*31*32/64*63*62 = 16 /21

IMO E
Three squares on a chessboard are chosen at random. The probability   [#permalink] 08 Oct 2019, 10:01
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