Question

with the solution show it and give the answer fast

Answer 1

Let S be sample space

From 100 - 999, (both are included)
No. of numbers = (999-100)+1= 899+1= 900

n(S)= no. of ways of choosing a no. from 900 no.s= 900C1= 900

E be the Event that a no. chosen from 100 to 999(both included),has at least one 0.
n(E)= no. of ways of choosing a no. from 100 to 999 such that it has at least one 0.

•°• The no. of zeroes can be 1 and 2, because,given [100 - 999], are 3 digit no.s.
(°•° All 3 zeroes is not a possibility)

The possible cases for Event E are:
•First let's see from 100 - 199
100,101,102,103,104,105,106,107,109,109,110,…120,130,140,150,160,170,180,190
•°• 100 - 199, write have 19 possibilities.
(100 - 199=> one hundred no.s)

For one hundred no.s------> 19 possibilities
For nine hundred no.s-----> 9*19=171 possibilities
This is nothing but n(E)
•°• n(E) = 171

Required probability = n(E)/n(S)= 171/900
This answer is in Option D

;)
hope it helps