Question

sum of the digits of a 5 digit number is 41 .find the probability that such a number is a divisible by 11​

Answer 1

Answer:

9+9+8+7+8

Step-by-step explanation:

9+9=18

18+8=26

26+7=33

33+8=41

Answer 2

Answer:

The correct  answer is 6/35

Step-by-step explanation:

To get the sum as 41, the following 5 digit combinations exist:

99995 →  5 digit numbers are 5, in this case

99986 → 5 digit numbers are 20, in this case

99977 → 5 digit numbers are 10, in this case

99887 → 5 digit numbers are 30, in this case

98888 → 5 digit numbers are 5, in this case

Therefore, 70  numbers exist  5+20+10+30+5= 70

Now we  have given that.  

For a 5 digit number of form (abcde) to be divisible by 11,

(a+c+e) - ( b+d) = 11, also (a+c+e) +(b+) = 41

Therefore,  a+c+e = 26, b+ = 15

Note that, (a+c+e) - (b+d) = 22 is not possible as it will give fractional values of (a+c+e) and (b+).

Also, (a+c+e) + (b+d) = 33 is not possible as it will give (a+c+e) = 37 and we know that, a,c, are digits.

(a,c,e) = (9,9,8) and (b, ) = (8,7) -------- (i)

or (a, c, e) = (9,9,8) and (b,) = (9,6) -------- (ii)

Therefore, After using the first equation we can make 6 numbers.

After, using second Equation we can make 6 numbers.

Therefore, the Number of favourable cases = 12.

Hence, the required probability will be:

12/70 = 6/35

which is the required answer:

to know more about the probability you may visit the below links.

https://brainly.in/textbook-solutions/q-complete-following-statements-probability-event-e-probability-1?source=quick-results#q-complete-following-statements-probability-event-e-probability

https://brainly.in/question/20137701