If two number cubes are tossed, what is the probability of getting a sum that is less than 6, given that 1 number cube shows a 3 ?

Assuming 6-sided number cubes.

Let the events
A=throwing less than 6
B=one of them is a three

Then P(A∩B)
=|{(1,3),(2,3),(3,1),(3,2)}| / 36
= 1/9

P(B)
=2/6
=1/3

Throwing less than six given one of them is a 3 is therefore the conditional probability of A given B, or
P(A|B)
=P(A∩B)/P(B)
=(1/9) /(1/3)
=1/3

To find the probability of getting a sum less than 6, given that one number cube shows a 3, we need to consider the possible outcomes.

When one number cube shows a 3, it means that we have two possibilities for the other number cube:
1. The other number cube shows a number from 1 to 2: In this case, the sum will be less than 6.
2. The other number cube shows a number from 4 to 6: In this case, the sum will be 7 or greater.

Therefore, we only need to consider the first possibility, where the other number cube shows a number from 1 to 2.

Out of the six possible numbers on a number cube, two numbers (1 and 2) satisfy this condition. So, the favorable outcomes are 2.

Since there are six equally likely outcomes for the other number cube, the total number of outcomes is 6.

The probability of getting a sum less than 6, given that one number cube shows a 3, is therefore:

Favorable outcomes / Total outcomes = 2 / 6 = 1/3.

Thus, the probability is 1/3.

To find the probability, we need to determine the number of favorable outcomes and the total number of possible outcomes.

First, let's consider the number cube that shows a 3. Since we know that one number cube shows a 3, we have fixed one of the outcomes.

Now, let's consider the other number cube. Since it is a number cube, it can show any number from 1 to 6. However, we are interested in finding the sum that is less than 6, which means the other number cube can only show numbers 1, 2, or 3.

So, the favorable outcomes are the combinations where the second number cube shows 1, 2, or 3. There are three favorable outcomes.

Now, let's consider the total number of possible outcomes. For the first number cube, we already know it shows a 3, so there is only one possibility. For the second number cube, there are six possibilities since it can show any number from 1 to 6.

Therefore, the total number of possible outcomes is 1 * 6 = 6.

To find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:

Probability = Number of Favorable Outcomes / Total Number of Possible Outcomes
= 3 / 6
= 1 / 2

So, the probability of getting a sum that is less than 6, given that one number cube shows a 3, is 1/2 or 0.5.

I have a different one so i don't know this one sorry!!