sixpointer wrote:
Your test souce is problematic - all of the questions have only four answers, and none of them have the correct answer.

1.Ticket numbered from 1 to 20 are mixed up and then a ticket is drawn at a random . What is the probablity that the ticket drawn has a number which is multiple of 3 or 5?
1/2,3/20,2/5,1/2

for this do we have to consider 15 also as it is divisible by 3 and 5 both ?
For such a low number, just count the number of wanted outcomes (those divisible by 3 or 5). Be sure to count each number only once - i.e. do not count 15 twice, since there's only one ticket with 15 on it:
multiples of 3: 3, 6, 9, 12, 15, 18 - total of 6 outcomes
multiples of 5: 5, 10, 15, 20 - total of 4 outcomes, (3 without '15', which is already counted on the other list)
Discount 15, the only member on both lists, so you have 6+3=9 wanted outcomes out of 20. None of the answers fit, but that is the correct answer.

2.In a box there are 8 red , 7 blue,and 6 green balls. One ball is picked up randomly . What is the probablity that the ball is niether red nor green?

2/3,3/4,7/19,8/21,9/21

Iam getting answer as 7/21

I agree.

3.In a class 30% student offered English 20% offered Hindi and 10% offered both . If a student is selected at random . What is the probablity that he has offered English or Hindi?

2/5,3/4,3/5,3/10
Same as first question - add the two groups, but discount the group belonging to both so as not to count it twice. The end formula for group A OR B = Group A+ group B - both

In this case, 30%+20%-10% = 40%