Question

Two dice are rolled . What is the probability that the sum of the faces will not 2 exceed 7? Given that at least one face shows a 4 ​

Answer 1

Answer:

We have got two dices here, under the assumption that each roll is fair.

Circumstances where at least one dice would be five would be to have the roll one of the two dices be five, or have two of them to roll a five.

The probability of {(one of them to roll a five ) + (two of them to roll a five)} would be (3/36), or (1/12),as there are 6*6*1 possible combinations of the numbers rolled, and among them, two correspond to one of the dices rolling a five, while one of them correspond to two dices rolling five.

The probability of getting dice rolls where sum of the numbers equal to seven would be (6/36), as the possible combinations are:

1 6

25

34

43

52

61

Adding the p values up, would net you the value of (1/4).

So the probability is (1/4).

Edit:As Dan Barker has said, 5 2 and 2 5 fall into both boundaries set by the question.

However, I also ignored the fact that there are actually more than three circumstances where one rolls a number five.

The possible circumstances are:

1 5

2 5

3 5

4 5

5 5

6 5

5 1

5 2

5 3

5 4

5 6.

Out of all of them, 5 2 and 2 5 correspond to one where the rolls add up to seven.

The probability , hence, is actually (11/36)+(6/36)-(2/36)=(5/12) This can give u a idea...

Step-by-step explanation: Pls mark me as brainliest