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Suppose four integers are chosen successively at random between 0 and 12, inclusive. Find the probability that:
a. they are all different?
b. not more than 2 are the same?

 Feb 18, 2020
 #1
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0,12=12

 

:)

 Feb 18, 2020
 #2
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I will give this a try!!

 

a - The first integer can be any integer between 0 and 12                      =13 / 13

      The probability that the 2nd number is different from the first           =12 / 13

      The probability that the 3rd number is different from the first two     =11 / 13

       The probability that the 4th number is different from the first three =10 / 13

Then, the overall probability that all four numbers are different from each other =13 /13  x  12 / 13  x  11 / 13  x  10 / 13 =17,160 / 13^4 =1,320 / 2,197 =60.08 %

 

b - [13 C 2  x  13^11] / 13^13 =6 / 13 =46.15 %

Note: Somebody should check this,please. Thanks.

 Feb 18, 2020

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