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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. $1/6$
Hello sir, why we have taken sample space 36 although we not throw dice again when we got 4,5,6 .

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
selected by
in favorable cases why  6 is considered ?
great explanation @Arjun.
IN THE PROBLEM HE SAID SUM MUST BE ATLEAT 6.....

THEN HOW DID U TAKE '6' AS ALONE SIR!!!!1 AS AN EVENT

If the value on the die is 1, 2, or 3, the die is rolled a second time.

So, if there are 4,5 or 6 on the first roll, there will not be any second roll. So, {6} is taken as a favorable case.

@arjun sir @bikram sir

I perfectly got this approach.

this question is given under Bayes' theorem in Go pdf, but what would be the events look like if try to do it Bayes' theorem. I've tried like this
P(E1)= Rolling a dice. P(E2)=getting 1/2/3 P(E3)= getting sum >=6 in second roll. but not able to continue, will you please help here.

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.
IN THE FINAL STEP.....

PLEASE EXPLAIN ME ELABORTE IN MANNER

The question is :-What is the probability that the sum total of values that turn up is at least 6?

so u may get 6 in first try also ..and it's Probability is 1/6.

NIce:))

So, this one is such intresting question:

If (1 or 2 or 3) appear than dice rolled second time and some cases are favorable, but if (4 or 5 or 6) appear than dice don't rolled out second time!

And now, the twist comes, they said total sum will be at least 6(it dosen't matter that you get it by 1 roll of dice or 2 roll of dice) so if 6 comes than we stop for rolling but it is our satisfying condition!

Such a great question!

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 .
+1 vote

P(getting sum is atleast 6) = P(getting 6 in the first roll of die ) OR P(getting 1,2 or 3 in first AND getting no  according  to result of first roll which gives sum atleast 6)

P(sum>=6) = 1/6 + (1/2*9/18)

=  5/12

@Rajesh,1/6 coz 6 can occur on either of d two dice.Isn't it?
@Devshree,

1/6 because 6 can come first time when we roll the dice,then we don't require to roll it again
Total Probablity = 1/6 + 3/6{9/18} = 5/12