Probability 2

For the five digits, there are totally 5P5 = 120 permutations.

a. For the number to be divisible by 2, then the last digit has to be '2' or '4'

We don't really need to consider the first 4 digits, just consider the probability of getting '2' or '4' as the last digit.

The required probability = 2/5


b. For the number to be divisible by 4, the last digit, of course, has to be ended by '2' or '4'.

But this time, we have to also consider the 2nd last digit. For the last digit to be '2', then the 2nd last digit has to be '1', '3' or '5'. (Because XXX12, XXX32, XXX52 is divisible by 4, but not XXX42).

For the last digit to be '4', then the 2nd last digit has to be '2'.

The probability = P(Getting '1', '3' or '5' as the 2nd last digit) X P('2' as the last digit) + P('2' as the 2nd last digit) X P('4' as the last digit)

= 3/5 X 1/4 + 1/5 X 1/4

= 1/5



2009-07-05 15:31:55 補充：
樓下朋友的答案一定錯，因為probability的數值不可能大過一

2009-07-05 22:07:44 補充：
002的答案錯，probability數值不能大於1

(a) if the digit is divisible by 2, the end digit is 2 or 4.
The probability (digit is divisible by 2) = 4! (2) = 24 x 2 = 48
(b) if the digit is divisible by 4, the end digit is 2 or . When the last digit is 2, the second last digit must be 1, 3 or 5. When the last digit is 4, the second must 2
The probability (digit is divisible by 4) = 3! (3)(1) + 3! (1)(1) = 24

2009-07-05 13:35:12 補充：
(a) the probability (digit is divisible by 2) = 4!(2)/5! = 2/5
(b) the probability (digit is divisible by 4) = (3!(3)(1)+3!(1)(1))/5! = 1/5