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geniusboy [140]
2 years ago
9

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

n even number?​
Mathematics
2 answers:
Alona [7]2 years ago
8 0

Answer:

0.444    (44.4%)

Step-by-step 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

zhuklara [117]2 years ago
6 0

Answer:

  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

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