gujacksongu6
24.07.2019 •
Mathematics
The number of three-digit numbers with distinct digits that be formed using the digits 1,2,3,5,8 and 9 is . the probability that both the first digit and the last digit of the three-digit number are even numbers .
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Ответ:
1.
2.
Ответ:
a)120
b)6.67%
Step-by-step explanation:
Given:
No. of digits given= 6
Digits given= 1,2,3,5,8,9
Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be
6P3= 6!/3!
= 6*5*4*3*2*1/3*2*1
=6*5*4
=120
Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:
As the first and last digits can only be even
then the form of number can be
a)2n8 or
b)8n2
where n can be 1,3,5 or 9
4*2=8
so there can be 8 three-digit numbers with both the first digit and the last digit even numbers
And probability = 8/120
= 0.0667
=6.67%
The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !
Ответ: