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

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)

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!

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

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)

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!

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

Posted from my mobile device