Conditional Probability - Part 1 Video Explanation

Dear team ,

In this video minute 6:18 the instructor tries to solve the questions “What is the probability that now the sum is 10, given that the first one is six?” and I kind of got confused with the answer.

The answer seems to be “… so the probability is now 1/6. The probability change from 1/12 to 1/6” but I can see in the whole probabilistic space that there is only 1 answer (6,4). Shouldn’t be then, the answer 1/36?

I’m trying to reason about it graphically and I’m reasoning that first P(A) P(all sum 100) = 3/36 (So only 3 of 36 can fit first condition ) then P(Start by 6 | All sum 10) evolves from the condition in such a way that among those 3 ( P(B) new probabilistic space) only 1 fits the second condition therefore the whole probabilistic space reduces to 1/36.

I get to understand it that P(A) = 1/12 and P(B) = 1/3 therefore P(B|A) = 1/12 * 1/3 = 1/36.

What do you think? could it be this a solution to the problem statement?

Hi @josetonyp!

First, welcome to our community! It is always good to see new people around here sharing their questions and thoughts! :smiley:

The answer would be \frac{1}{36} if the question asked was

What is the probability of, when throwing two dice, the first one lands in 6 and the second one lands in 4?

However, the question is slightly different. It is now a conditional probability.

“What is the probability that now the sum is 10, given that the first one is six?”

Note that we already know that the first dice is six, so in this case, there is no random event anymore related to this particular dice. Its value were already determined. So it turns out that the only random event to occur is the throwing of the next dice. And in this case, there is only one side that sums 10, i.e., 4. Therefore there is a \frac{1}{6} chance of landing in 4, thus summing 10 with the previous dice.

If you still can’t get it, please let me know and we can work it together.

Cheers,
Lucas

1 Like

Hi @lucas.coutinho ,

Thanks a lot for you answer, I miss understood the problem greatly looking for a more complex solution. Thanks for adding this perspective.

Best,
Jose