A multiple-choice quiz
by rodney_indy.
Estimated time: 4 mins.

Scroll down to the bottom for the answer key.

Most Recent Scores

Sep 13 2023
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Guest 136: 0/10Sep 11 2023 : Guest 163: 2/10

Sep 07 2023 : Guest 175: 2/10

Sep 06 2023 : Guest 199: 3/10

Sep 05 2023 : Guest 1: 3/10

Sep 04 2023 : Guest 102: 7/10

Sep 02 2023 : Guest 68: 6/10

Aug 21 2023 : Guest 97:

Aug 20 2023 : Guest 110: 4/10

Score Distribution

Quiz Answer Key and Fun Facts

Answer:
**1/36**

There are 6 possibilities for the first die and 6 possibilities for the second die. Hence there are 6 * 6 = 36 possible dice rolls. There is only one of them: (2,5) which has 2 on the first die and 5 on the second die. Thus the probability is 1/36.

There are 6 possibilities for the first die and 6 possibilities for the second die. Hence there are 6 * 6 = 36 possible dice rolls. There is only one of them: (2,5) which has 2 on the first die and 5 on the second die. Thus the probability is 1/36.

Answer:
**1/18**

This event consists of the outcomes (2,6), (6,2). Hence it has probability 2/36 = 1/18.

This event consists of the outcomes (2,6), (6,2). Hence it has probability 2/36 = 1/18.

Answer:
**11/36**

The probability that the first die is 3 is 1/6, the probability that the second die is 3 is 1/6, and the probability that both dice are 3 is 1/36. By the principle of inclusion/exclusion, the probability that the first die or the second die is 3 is just 1/6 + 1/6 - 1/36 = 11/36. This probability is useful in backgammon - this is the probability of entering off the bar when there is only one open point in the opponent's table.

The probability that the first die is 3 is 1/6, the probability that the second die is 3 is 1/6, and the probability that both dice are 3 is 1/36. By the principle of inclusion/exclusion, the probability that the first die or the second die is 3 is just 1/6 + 1/6 - 1/36 = 11/36. This probability is useful in backgammon - this is the probability of entering off the bar when there is only one open point in the opponent's table.

Answer:
**5/36**

The outcomes in the event "the sum is 8" are: (2,6), (3,5), (4,4), (5,3), and (6,2). Since there are 5 ways the sum can be 8, the probability is 5/36.

The outcomes in the event "the sum is 8" are: (2,6), (3,5), (4,4), (5,3), and (6,2). Since there are 5 ways the sum can be 8, the probability is 5/36.

Answer:
**1/6**

This is a conditional probability question. We are given that the first die was 3: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

Of these 6 possibilities, (3,5) is the only outcome where the dice sum to 8. So the probability is 1/6.

This is a conditional probability question. We are given that the first die was 3: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

Of these 6 possibilities, (3,5) is the only outcome where the dice sum to 8. So the probability is 1/6.

Answer:
**1/5**

This is another conditional probability question. We are given that the dice sum to 8: (2,6), (3,5), (4,4), (5,3), (6,2)

Of the 5 ways that the dice sum to 8, only one has the first die 3: (3,5). Hence the probability is 1/5.

This is another conditional probability question. We are given that the dice sum to 8: (2,6), (3,5), (4,4), (5,3), (6,2)

Of the 5 ways that the dice sum to 8, only one has the first die 3: (3,5). Hence the probability is 1/5.

Answer:
**5/6**

The ways in which a 7 can be rolled are: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 ways a sum of 7 can be rolled. There are 36 total dice rolls, thus there are 30 ways a sum of 7 cannot be rolled. Hence the probability is 30/36 = 5/6.

The ways in which a 7 can be rolled are: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 ways a sum of 7 can be rolled. There are 36 total dice rolls, thus there are 30 ways a sum of 7 cannot be rolled. Hence the probability is 30/36 = 5/6.

Answer:
**5/6**

We are given that the sum is not 7. In question number 7, we saw that there were 30 ways the sum could not be 7. The ways we can roll a sum of 8 are: (2,6), (3,5), (4,4), (5,3), and (6,2). So among the 30 ways we cannot roll a sum of 7, there are 25 ways we cannot roll a sum of 8. Hence the probability is 25/30 = 5/6.

We are given that the sum is not 7. In question number 7, we saw that there were 30 ways the sum could not be 7. The ways we can roll a sum of 8 are: (2,6), (3,5), (4,4), (5,3), and (6,2). So among the 30 ways we cannot roll a sum of 7, there are 25 ways we cannot roll a sum of 8. Hence the probability is 25/30 = 5/6.

Answer:
**5/6**

Let E be the event "the first die is a 5" and F be the event "the second die is a 2." E and F are independent events. Hence so are the complements of E and F. This means that the probability of the complement of E given the complement of F is just the probability of the complement of E, which is 1 - 1/6 = 5/6.

Let E be the event "the first die is a 5" and F be the event "the second die is a 2." E and F are independent events. Hence so are the complements of E and F. This means that the probability of the complement of E given the complement of F is just the probability of the complement of E, which is 1 - 1/6 = 5/6.

Answer:
**64/1323**

Let x = the probability of rolling a 6. Then 2x is the probability of rolling a 5, 2*2x = 4x = the probability of rolling a 4, 8x = the probability of rolling a 3, 16x = the probability of rolling a 2, and 32x = the probability of rolling a 1. These are all the outcomes, so these probabilities must sum to 1: x + 2x + 4x + 8x + 16x + 32x = 1 which simplifies to 63x = 1 which means x = 1/63. So Pr(1) = 32/63, Pr(2) = 16/63, Pr(3) = 8/63, Pr(4) = 4/63, Pr(5) = 2/63, and Pr(6) = 1/63. There are six ways to roll a sum of 7:

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

By independence, their corresponding probabilities are

(32/63)*(1/63), (16/63)*(2/63), (8/63)*(4/63), (4/63)*(8/63), (2/63)*(16/63), and (1/63)*(32/63).

Each probability is 32/63^2. We add up these six probabilities to get the probability of rolling a sum of 7: 6*(32/63^2) = 64/1323.

I hope you enjoyed my quiz! Thanks for playing!

Let x = the probability of rolling a 6. Then 2x is the probability of rolling a 5, 2*2x = 4x = the probability of rolling a 4, 8x = the probability of rolling a 3, 16x = the probability of rolling a 2, and 32x = the probability of rolling a 1. These are all the outcomes, so these probabilities must sum to 1: x + 2x + 4x + 8x + 16x + 32x = 1 which simplifies to 63x = 1 which means x = 1/63. So Pr(1) = 32/63, Pr(2) = 16/63, Pr(3) = 8/63, Pr(4) = 4/63, Pr(5) = 2/63, and Pr(6) = 1/63. There are six ways to roll a sum of 7:

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

By independence, their corresponding probabilities are

(32/63)*(1/63), (16/63)*(2/63), (8/63)*(4/63), (4/63)*(8/63), (2/63)*(16/63), and (1/63)*(32/63).

Each probability is 32/63^2. We add up these six probabilities to get the probability of rolling a sum of 7: 6*(32/63^2) = 64/1323.

I hope you enjoyed my quiz! Thanks for playing!

This quiz was reviewed by FunTrivia editor crisw before going online.

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