Question

Suppose a 5-digit number is formed using the digits from 1 to 9 (without replacement). What is the probability that it will be an even number?​

Answer 1

0.444 (44.4%)

Explanation:

All possible ending: 1,2,3,4,5,6,7,8,9 ... 9

ending with 2,4 6,8 to make even number: 4

___ ___ ___ ___ ___

even number : 8 * 7 * 6 * 5 * 4

All 5 digit without repeating: 8 * 7 * 6 * 5 * 9

possibility = (8*7*6*5*4) / (8*7*6*5*9)

= 4/9

= 0.444 (44.4%)

_______________________________________________

(4 * ₈P₄) / (9 * ₈P₄) = 4/9

Answer 2

1. First take 5 empty. Digits.
2. The no of digits are 9 (1-9).
3. The last no must be even so . The total no of even no’s . Are 4 (2,4,6,8)
4. The probability of last digit is 4
5. There are remaing 8 digits so. Place them where. U required. The probability are. 8, 7,6,5(no reputations)
6. The final answer is 4*8*7*6*5=6720 ways