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Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?

1. $10/21$
2. $5/12$
3. $2/3$
4. $1/6$

Here our sample space consists of 3 + 3 * 6 = 21 events- (4), (5), (6), (1,1), (1,2) ... (3,6).

Favorable cases = (6), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3,5), (3,6)

Required Probability = No. of favorable cases/Total cases = 10/21

But this is wrong way of doing. Because due to 2 tosses for some and 1 for some, individual probabilities are not the same. i.e., while (6) has 1/6 probability of occurrence, (1,5) has only 1/36 probability. So, our required probability

= 1/6 + (9 * 1/36) = 5/12
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in favorable cases why  6 is considered ?
great explanation @Arjun.
P(getting a 1)=1/6 with 1 you can form 6,7 so p(sum>=6)=2/6

P(getting a 2)=1/6 with 2 you can form 6,7,8 so p(sum>=6)=3/6

P(getting a 3)=1/6 with 3 you can form 6,7,8,9 so p(sum>=6)=4/6

now every time we are going to form a number we require a particular number whose probability is 1/6

so total probability =1/6(2/6+3/6+4/6+1)=15/36=5/12
please mention bayes' theorem and the fact that these events are independent

also mention the case where you get 6 on you first roll

No , its ok .

last one is added for anything that is not 1 or 2 or 3 , right .

Following is the wrong approach which I had done initially and max. people think like this...So, we should know this(thanks @BILLY for identifying the mistake...) ///The last img is the correct approach..

And the correct approach goes like below...

edited
@rajesh plzz elaborate how 24 ways ?
Thanks @BILLY and Plz see the edited ans.