An experiment is conducted where a dice is rolled following the rules given below:
It is rolled only once if the first dice throw turns up to 1, 2 or 3. It is rolled twice if the first throw turns up 4, 5 or 6. What is the probability that the sum of all values that turn up in the above experiment is >= 6?
In my opinion, the problem is about reducing the sample space, i.e. considering the cases which are favorable. Once we are able to reduce the sample size, we can get to the final answer.
I have tried doing this question from the conventional approach of directly finding the required sample space. As per me, S = {1, 2, 3,41,42,43,44,45,46,51,52,53,54,55,56,61,62,63,64,65,66} Concluding from the above sample space, P(sum>=6) = 17/21, but the actual answer is marked as 5/12. I was thinking of another approach, by doing the question using the multiply method. For the required case, first we have 3/6 probability(i.e. 4,5, or 6) and then I am confused, what should be the next element that should be multiplied.
I am not sure why the answer is marked as 5/12. Is the answer given wrong or I am making some mistake?
Let’s break down the problem:
If the first roll is 1, 2, or 3, the dice is rolled only once. The sum of the values can never be greater than or equal to 6. So, the probability in this case is 0.
If the first roll is 4, 5, or 6, the dice is rolled twice. We need to find the probability that the sum of the two rolls is greater than or equal to 6.
For a sum of 6: The possible outcomes are (1,5), (2,4), (3,3), (3,3), (4,2), (5,1) each with a probability of 1/36. So, the total probability is 6/36.
For a sum of 7: The possible outcomes are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) each with a probability of 1/36. So, the total probability is 6/36.
For a sum of 8: The possible outcomes are (2,6), (3,5), (4,4), (4,4), (5,3), (6,2) each with a probability of 1/36. So, the total probability is 6/36.
For a sum of 9: The possible outcomes are (3,6), (4,5), (5,4), (6,3) each with a probability of 1/36. So, the total probability is 4/36
For a sum of 10: The possible outcomes are (4,6), (5,5), (5,5), (6,4) each with a probability of 1/36. So, the total probability is 4/36
For a sum of 11: The possible outcomes are (5,6), (6,5) each with a probability of 1/36. So, the total probability is 2/36
For a sum of 12: The possible outcome is (6,6), (6,6) each with a probability of 1/36. So, the total probability is 2/36.
So, the total probability for the sum to be greater than or equal to 6 when the dice is rolled twice is (6+6+6+4+4+2+2)/36 = 30/36 = 5/6.
Since the first roll can be 4, 5, or 6 with a probability of 1/2, the overall probability is, (5/6) * (1/2) = 5/12. This is your answer.