Question

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 .

Answer 1

1.

`6\cdot5\cdot4=120`

2.

`|\Omega|=120\\|A|=2\cdot4\cdot1=8\\\\P(A)=(8)/(120)=(1)/(15)\approx6.7\%`

Answer 2

a)120

b)6.67%

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% !