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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. $1/6$
asked in Probability by Veteran (12.9k points)   | 1.1k views

4 Answers

+15 votes
Best answer
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
answered by Veteran (281k points)  
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
+7 votes

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...

answered by Veteran (16.3k points)  
edited by
@rajesh plzz elaborate how 24 ways ?
Thanks @BILLY and Plz see the edited ans.
IN THE FINAL STEP.....

WHY DID U ADD 1*(1/6)?

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!

+5 votes
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
answered by Veteran (13k points)  
please mention bayes' theorem and the fact that these events are independent

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

your answer is correct but explaination is incomplete
No , its ok .

 

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

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

answered by (29 points)  
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