If you roll two fair six sided dice what is the, №9844911, 26.10.2020 13:11

Mathematics

If you roll two fair six-sided dice, what is the probability that a least one dice shows a 3?

Step-by-step answer

P Answered by Master

5

11/36

Step-by-step explanation:

So here’s the probability of rolling one three or more on one dice:

2,3

1,3. 5 chance of at least one 3 on this dice

4,3

5,3

6,3

On other die:

3,2

3,4

3,1

3,5

3,6. 6 chance of rolling at least one 3 on this dice

3,3

5+6=11

11/36 possible combinations contain at least one three

Hope this helps!

Mathematics

If you roll two fair six-sided dice, what is the probability that at least one die shows a 3?

Step-by-step answer

P Answered by PhD

12

-- Two fair dice can land in 36 possible ways. (6 ways each).

-- The successful tosses are

      3 (not 3)
   (not 3) 3
and    3 3

Three of the 36 possible outcomes are successful.

-- The probability of a successful toss is

   (3) / (36) = 1/12 = (8 and 1/3) % .

Mathematics

If you roll two fair six-sided dice, what is the probability that at least one die shows a 3?

Step-by-step answer

P Answered by Specialist

9

11/36

Step-by-step explanation:

khan acadamy

Mathematics

If you roll two fair six-sided dice, what is the probability that at least one die shows 3?

Step-by-step answer

P Answered by Specialist

23

P(3) = 1/12

Step-by-step explanation:

12 is the denominator because you have two die with six sides.

Mathematics

If two standard six-sided dice are tossed, what is the probability that a 5 is rolled on at least one of the two dice? express your answer as a common fraction.

Step-by-step answer

P Answered by PhD

We define the probability of a particular event occurring as:
$\frac{number\ of \ desired\ outcomes}{number\ of\ possible\ outcomes}$

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are independent, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for each of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.

Now that we've found the number of possible outcomes, we need to find the number of desired outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is at least one 5 rolled. It turns out, there are only 3:

(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5

So, we have $\frac{3}{36} = \frac{1}{12}$ probability of rolling at least one 5.

Mathematics

If two standard six-sided dice are tossed, what is the probability that a 5 is rolled on at least one of the two dice? express your answer as a common fraction.

Step-by-step answer

P Answered by PhD

We define the probability of a particular event occurring as:
$\frac{number\ of \ desired\ outcomes}{number\ of\ possible\ outcomes}$

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are independent, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for each of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.

Now that we've found the number of possible outcomes, we need to find the number of desired outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is at least one 5 rolled. It turns out, there are only 3:

(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5

So, we have $\frac{3}{36} = \frac{1}{12}$ probability of rolling at least one 5.

Mathematics

Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)

Step-by-step answer

P Answered by Master

The probability that there is at least one pair but not a three of a kind is 0.6944.

Step-by-step explanation:

We have the case when we have five dices and we roll them and we need that there is at least one pair but not a three of a king which is Two dice showing the same value, but no three dice showing the value.

Total no of outcomes = 6^5 = 7776

We will consider the other cases and then subtract them from the total outcomes.

No of outcomes of 5 of a kind = 6

No of outcomes 4 of a kind and 1 different = (6*5)*5 = 150

No of outcomes with 3 of a kind and 1 different pairs = (6*1*1*5*1)*5C2= 300

No of outcomes with 3 of a kind and two different = 6*5*4*10= 120*10=1200

No of outcomes when all are different = 6! = 720

These all the cases when subtracted from the total will represent the total cases in which there are two dice showing the same value.

Total cases = 6+150+300+1200+720

= 2376

Required no of outcomes = 7776-2376 = 5400

Probability = $\frac{5400}{7776}$ = 0.6944

Therefore the probability for the case of at least one pair but not a three of a kind is 0.6944.

Mathematics

Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)

Step-by-step answer

P Answered by Specialist

The probability that there is at least one pair but not a three of a kind is 0.6944.

Step-by-step explanation:

We have the case when we have five dices and we roll them and we need that there is at least one pair but not a three of a king which is Two dice showing the same value, but no three dice showing the value.

Total no of outcomes = 6^5 = 7776

We will consider the other cases and then subtract them from the total outcomes.

No of outcomes of 5 of a kind = 6

No of outcomes 4 of a kind and 1 different = (6*5)*5 = 150

No of outcomes with 3 of a kind and 1 different pairs = (6*1*1*5*1)*5C2= 300

No of outcomes with 3 of a kind and two different = 6*5*4*10= 120*10=1200

No of outcomes when all are different = 6! = 720

These all the cases when subtracted from the total will represent the total cases in which there are two dice showing the same value.

Total cases = 6+150+300+1200+720

= 2376

Required no of outcomes = 7776-2376 = 5400

Probability = $\frac{5400}{7776}$ = 0.6944

Therefore the probability for the case of at least one pair but not a three of a kind is 0.6944.

Mathematics

1. what is the median of the following set of numbers? 48.4, 47, 43, 47, 43.6, 41.6, 45, 51.4, 38.6 hint: order the numbers from least to greatest first! a. 48.4 b. 43.6 c. 45 2. each of the five math teachers are comparing the number of students in their homeroom classes. the number of boys and girls in each teachers’ classes are recorded in the table below. hint: mean is the average. which statement best describes the means of the two sets of data? a. the mean number of girls in 7 times greater than the mean numbers of boys in the homeroom classes.

Step-by-step answer

P Answered by Specialist

9

1. C, 45.

2. I choose B. I'm not sure if that's correct though.

3. I choose B as well.

4. I choose D.

5. A, 3/4

6. D, 1/12

7. C, 1/9

8. Part A. 80% Chance. Part B. 6% Chance. Part C. 16.7% Chance.

2, 3, 4 I'm not sure if their correct but that's what I got. I really hopes this helps!

Mathematics

1. what is the median of the following set of numbers? 48.4, 47, 43, 47, 43.6, 41.6, 45, 51.4, 38.6 hint: order the numbers from least to greatest first! a. 48.4 b. 43.6 c. 45 2. each of the five math teachers are comparing the number of students in their homeroom classes. the number of boys and girls in each teachers’ classes are recorded in the table below. hint: mean is the average. which statement best describes the means of the two sets of data? a. the mean number of girls in 7 times greater than the mean numbers of boys in the homeroom classes.

Step-by-step answer

P Answered by Master

9

1. C, 45.

2. I choose B. I'm not sure if that's correct though.

3. I choose B as well.

4. I choose D.

5. A, 3/4

6. D, 1/12

7. C, 1/9

8. Part A. 80% Chance. Part B. 6% Chance. Part C. 16.7% Chance.

2, 3, 4 I'm not sure if their correct but that's what I got. I really hopes this helps!

If you roll two fair six sided dice what is the probability that at least one dice shows a three

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