In a certain parking lot, 31 percent of the cars are black, 20 percent

Question

In a certain parking lot, 31 percent of the cars are black, 20 percent are white, 16 percent are gray, and 13 percent are silver. What is the probability that if 4 cars are randomly selected, exactly 2 will not be one of the above colors?

A. 16/625
B. 16/125
C. 96/625
D. 8/25
E. 1/2

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Coolsandyp51 · Answer

There are 4 cars and exactly 2 be different.Thus, 2 different and 2 of above gives 4!/2!*2! combinations.Now probability of other car is 1/5 and probability of above mentioned car is 4/5. Thus total probability is (4!/2!*2!)*1/5*1/5*4/5*4/5==96/625

lasttime · Answer

Suppose there are 10 cars. 8 cars have defined colours, whereas, 2 cars have random other colours.

Probability of choosing any of the defined colour cars = 8/10
Probability of choosing one of the other coloured car = 1/10

1/10*1/10*8/10*8/10*4! = 96/625

mariaraashidkoul · Answer

Shouldn't the probability of choosing other colored car be 2/10?

mariaraashidkoul · Answer

why have you divided by 2! twice? Shouldn't it be just once?

vaibhav1221 · Answer

Imagine you need to arrange 4 letters in any random order. In how many ways can you do it? 
4! Ways.
Now imagine if these letters are “a, a, b, b”
Now the total number of ways cannot be 4!. Why? Because the method by which we calculate assumes that all the letters are different. So to accommodate that you need to divide 4! by factorial of the number of times each letter is being repeated. Here “a” comes twice and “b” comes twice so, we need to divide by 2!*2!

In this question because, a = 1/5 (probability of selecting the undefined colour car)
b = 4/5 (probability of selecting the defined colour car)

So, probability = (1/5*1/5*4/5*4/5)*4!/2!2!

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vritti_s · Answer

why are we multiplying 1/5 and 4/5 two times?

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In a certain parking lot, 31 percent of the cars are black, 20 percent

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