Question

Marti rolls two dice what is the probability that the sum of the dice is 7 given that the first die is 2

Answer 1

Answer:

The probability such that the sum of the dice is 7 given that the first die is 2 is:

1/36

Step-by-step explanation:

The sample space that is obtained by rolling two die is given as:

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

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

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

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

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

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

The outcomes such that the the sum of the dice is 7 given that the first die is 2 is:

(2,5)

Total outcomes=36.

Number of favorable outcomes=1.

Hence, the probability is defined as:

Probability= Number of Favorable outcomes/Total number of outcomes

Hence, Probability=1/36

Answer 2

You can use the product rule(also called chain rule) of probability to get the result.

Thus, the probability that the sum of the dice is 7 given that the first die is 2 is $P(A|B) = \dfrac{1}{6}$

What is product rule of probability?

If A and B are two events, then by product rule of probability, we have:

$P(A \cap B) = P(A) P(B|A) = P(B)P(A|B)$

How to find the needed probability?

Let the event of getting a sum of 7 in a roll of two dice is  denoted by A

Let the event of getting 2 on first die is denoted by B

Then we have:
$P(A) = \dfrac{\text{number of favorable case}}{\text{number of total cases}} \\\\ P(A) = \dfrac{count( (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) )}{6 \times 6} = \dfrac{6}{36} = \dfrac{1}{6}$

And

$P(B) = \dfrac{\text{number of favorable case}}{\text{number of total cases}} \\\\ P(B) = \dfrac{1}{6}$(since we don't care what comes on second die and thus the probability of getting 2 on first die depends only on first die and its probability is 1 over 6 because of six equally probable numbers to come on it)

And we have:

$P(A \cap B) = \dfrac{\text{number of favorable case}}{\text{number of total cases}} \\\\P(A \cap B) = \dfrac{count( (2,5) )}{6 \times 6} = \dfrac{1}{36}$sum as 7 and first die having 2 is possible only when second die gets 5 on it and this only occurs once in total 36 possible outcomes)  

The probability that the sum of the dice is 7 given that the first die is 2 is $P(A|B)$

Thus, using product rule of probability here:

$P(A \cap B) = P(B)P(A|B)\\\\ \dfrac{1}{36} = \dfrac{1}{6}P(A|B)\\\\ P(A|B) = \dfrac{6}{36} = \dfrac{1}{6}$

Thus, the probability that the sum of the dice is 7 given that the first die is 2 is $P(A|B) = \dfrac{1}{6}$

Learn more about product rule of probability here:

https://brainly.com/question/3932712