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

Question

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

firas92 · Accepted Answer

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}

Answer is E

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chetan2u · Answer

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

htmtruong · Answer

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

chetan2u · Answer

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!

Kinshook · Answer

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

Kritisood · Answer

Hi for the first approach first we selected 32C2*32C1 now we have to arrange. We have to arrange three colors out of which two are the same so why don't we multiple with 3!/2! (RRW)

KarishmaB · Answer

There are various ways of choosing 2 of one colour and 1 of the other.

Take one case:
BBW - (32/64) * (31/63) * (32/62)

But this can be arranged in 3 ways (BBW, BWB, WBB) 
and in another 3 ways with white taken twice (WWB, WBW, BWW)

Total Probability = (32/64) * (31/63) * (32/62) * 3 * 2 = (16/21)

Answer (E)

CrackverbalGMAT · Answer

There are a total of 64 squares on a chessboard of which 32 are black and the remaining 32 are white.
 
If three squares are to be chosen at random, the total number of ways to do this task =  64_C_3  =  \frac{64 * 63 * 62 }{ 3*2*1} . This can be simplified one step further and written as 64 * 21*31. Do not try to evaluate the exact value of the above product.

We are trying to find the probability that, of the 3 squares we selected, 2 are of one color and the remaining of the other color. This means, there can be two cases.

Case 1:  Select 2 black squares and 1 white square.
This can be done in  32_C_2  *  32_C_1  ways. 
 32_C_2 * 32_C_1  =  \frac{32 * 31 }{ 2}  * 32 = 32*31* 16.

Case 2: Select 2 white squares and 1 black square.
This can be done in 32_C_2 * 32_C_1 ways too which means, 32*31*16 ways.

Therefore, total number of favourable outcomes = 32*31*16 + 32*31*16 = 32*31*16*2 = 64*31*16.

Probability =  \frac{Number of favourable outcomes }{ Total possible outcomes}  =  \frac{64*31*16 }{ 64*31*21} , which simplifies to give us  \frac{16}{21} .

The correct answer option is E .

Hope that helps!

Fdambro294 · Answer

Number of favorable outcomes:

Scenario 1: 2 black and 1 white

(32 c 2) (32 c 1)

Or

Scenario 2: 2 white and 1 black

(32 c 2) (32 c 1)

Total possible outcomes:

All 64 squares — how many different ways to choose 3

(64 c 3)

Answer:

(2) (32 c 2) (32 c 1)
________________
(64 c 3)

Reduced: answer E

16/21

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Jxmes · Answer

why is this set up as a 'without replacement'? you're not removing the chess squares from the chessboard, right? each choice should be independent of the other.

Bunuel · Answer

The denominator reduces there because we cannot choose the same square again.

Jxmes · Answer

fair enough - i guess what i was getting at is that the question is a bit ambiguous. one could interpret each choice as being independent of the other, like coin flips, which would lead to 3/4 as the answer i believe. if the question can be edited, i'd suggest adding 'different' before squares.

perhaps i am the one missing something though, and it is somehow implied.

Bunuel

The denominator reduces there because we cannot choose the same square again.

perhaps i am the one missing something though, and it is somehow implied.

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

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

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