A die is rolled twice. What is the probability that the sum of the rolls is less than 4 given that one of the rolls is a 1?

The answer is 3/11. Could someone tell me how to get this. Thanks?

Possibilities with rolling 3 on ONE of the rolls:

First roll: 1,1 | 1,2 | 1,3 | 1,4 | 1,5 | 1,6
Second roll: 1,1 | 2,1 | 3,1 | 4,1 | 5,1 | 6,1

Eliminate 1,1 from second roll because it is the same as in the first, therefore there are 11 possibilities with 1 as either the first or second roll.

Find the rolls that are less than 4.
(1,1) (1,2) (2,1). 3 rolls are less than 4.

Therefore P(x < 4) = 3 / 11.

Sure, let me calculate that for you... just kidding! Let's use humor to lighten up the math a bit.

Well, if you have a die that rolls every possibility, it's like your decision-making skills—very random! We're looking for the probability that the sum of the rolls is less than 4 given that one of the rolls is a 1. So, let's break it down.

In a classic dice throw, you can have six possible outcomes for each roll. But we already know that one of the rolls is a 1, so let's focus on the second roll.

If the first roll is a 1, we are left with five possibilities for the second roll: 1, 2, 3, 4, or 5. Out of these five, only one combination yields a sum less than 4, and that is 1 + 1.

So, the probability of getting a sum less than 4, given that one of the rolls is a 1, is 1 out of 5 possibilities.

However, that's only considering the case where the first roll is a 1. But since the probability of rolling a 1 on one throw is 1/6, we need to take that into account too.

So, the overall probability would be (1/6) x (1/5) = 1/30.

However, we need to consider the case where the second roll is a 1. In this case, the possibilities are 2, 3, 4, 5, or 6 for the first roll. Out of these five possibilities, two combinations yield a sum less than 4: 1 + 2 and 1 + 3.

So, the probability of getting a sum less than 4 given that the second roll is a 1 is 2 out of 5 possibilities.

Again, we need to take into account the probability of rolling a 1 on the second throw, which is also 1/6.

So, the overall probability would be (5/6) x (1/6) x (2/5) = 1/30.

Now, we can add both probabilities together: 1/30 + 1/30 = 2/30 = 1/15.

But wait, we're not done yet! We need to subtract the probability of both rolls being 1, as we counted that twice.

The probability of both rolls being 1 is (1/6) x (1/6) = 1/36.

So, the final probability is 1/15 - 1/36 = 3/36 - 1/36 = 2/36 = 1/18.

Voila! The probability that the sum of the rolls is less than 4, given that one of the rolls is a 1, is 1/18.

Well, math can be a bit dicey, but with a little humor, it becomes a little more bearable, don't you think? Happy calculating!

To find the probability that the sum of the rolls is less than 4 given that one of the rolls is a 1, we need to consider all the possible outcomes. There are 6 possible outcomes when rolling a die twice: (1,1), (1,2), (1,3), (1,4), (1,5), and (1,6).

Out of these 6 outcomes, only 3 of them have a sum less than 4: (1,1), (1,2), and (1,3).

Therefore, the probability that the sum of the rolls is less than 4 given that one of the rolls is a 1 is 3/6, which simplifies to 1/2.

However, we need to consider that we are given that one of the rolls is a 1. So, out of the 6 possible outcomes mentioned above, only 2 of them have a 1 in the second roll: (1,1) and (1,2).

Therefore, the probability that the sum of the rolls is less than 4 given that one of the rolls is a 1 is 2/6, which simplifies to 1/3.

However, since we are asked for the probability, given that one of the rolls is a 1, the total possible outcomes are now reduced. Instead of 6 outcomes, we only have 5 possible outcomes: (1,1), (1,2), (1,3), (1,4), (1,5).

Out of these 5 outcomes, only 2 of them have a sum less than 4: (1,1), and (1,2).

Therefore, the probability that the sum of the rolls is less than 4 given that one of the rolls is a 1 is 2/5.

Hence, the answer is 2/5 or 0.4. It is not 3/11.

To solve this problem, we can use conditional probability. Let's break down the steps to calculate the probability.

Step 1: Determine the sample space.
When a die is rolled twice, there are 6 possible outcomes for each roll, resulting in a total of 6 * 6 = 36 possible combinations. These 36 combinations represent our sample space.

Step 2: Identify the favorable outcomes.
We know that one of the rolls is a 1. So, let's determine the possible combinations when at least one of the rolls is a 1:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)

There are 11 favorable outcomes in this case.

Step 3: Calculate the probability.
The probability of an event occurring is calculated by dividing the number of favorable outcomes by the total number of outcomes (sample space).

P(sum < 4 | one roll is a 1) = Favorable outcomes / Total outcomes
= 11/36

Therefore, the probability that the sum of the rolls is less than 4 given that one of the rolls is a 1 is 11/36.

Note: If we simplify the fraction, we get 3/11, which is equivalent to 11/36.

The rolls are independent. If you have already obtained a 1, you are asking for the probability of getting either a 1 or 2 to add to "less than 4."

1/6 + 1/6 = 2/6 = 1/3

With 6 possibilities for any roll of a die, there is no way of getting a denominator of 11.

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