# Week 1 - Summative Quiz - Q2

Case 1: If the initial roll is less than 4, you roll two dice and sum the results.

In this case, the possible outcomes that result in a sum of 6 are:

1 + 5 = 6
2 + 4 = 6
3 + 3 = 6
4 + 2 = 6
5 + 1 = 6

The total number of outcomes in this case is 5 * 6 = 30 (since each dice has 6 sides). However, we only consider the outcomes where the sum is 6, so the probability in this case is 5/30 = 1/6.

Case 2: If the initial roll is greater than 4, you roll only one dice and use the result.

In this case, the possible outcome that results in a final result of 6 is rolling a 6 on the dice.

The total number of outcomes in this case is 6 (since each dice has 6 sides), and only one of those outcomes results in a final result of 6. Therefore, the probability in this case is 1/6.

Now, we need to calculate the probability of each case occurring.

The probability of the initial roll being less than 4 is 3/6 (since there are three possible outcomes: 1, 2, and 3).

The probability of the initial roll being greater than 4 is 3/6 (since there are three possible outcomes: 5, 6).

Finally, we calculate the overall probability by considering both cases:

P(final result of 6) = P(less than 4) * P(sum of 6 | less than 4) + P(greater than 4) * P(result of 6 | greater than 4)
= (3/6) * (1/6) + (3/6) * (1/6)
= 3/36 + 3/36
= 6/36
= 1/6

Therefore, the probability of getting a final result of 6 after this experiment is 1/6 or approximately 0.1667.
Why result <> 1/6? Can anyone explain for me

two dices have 6*6 out comes so case 1 is 5/36

Hello @Phu_Nguyen ,

Though it’s very unclear, but apparently this has lot of contradictions from line to line and multiple mistakes.

As a mentor, I shouldn’t be saying the answer.
I would suggest you to revise Probability from the very basics.

Happy learning!!

With regards,
Nilosree Sengupta

@nilosreesengupta please correct me if I am wrong,

P(final result of 6) = P(sum of 6 | 2-dice game) + P(result of 6 | 1-dice game)

Per the lecture: P(A|B) = P(A,B) / P(B), hence

P(sum of 6 | 2-dice game) = P(sum of 6 with 2 dice) / P(2-dice game)
P(result of 6 | 1-dice game) = P(sum of 6 with 1 dice) / P(1-dice game)

This is the formula logic but it does not result in any of the possible choices.

On the other hand, I believe the right answer to be:
P(final result of 6) = P(sum of 6 with 2 dice) + P(sum of 6 with 1 dice)

where,
P(sum of 6 with 2 dice) = P(2-dice game) * P(sum of 6 | 2-dice game)
P(sum of 6 with 1 dice) = P(1-dice game) * P(result of 6 | 1-dice game)

Which result is also not included in the possible choices (maybe a typo?)