What is the probability that the sum of two dice will be greater than or equal to 10 given that the first die is a 5?

Next section: Probability of A and B A conditional probability is the probability of an event given that another event has occurred. For example, what is the probability that the total of two dice will be greater than 8 given that the first die is a 6? This can be computed by considering only outcomes for which the first die is a 6. Then, determine the proportion of these outcomes that total more than 8. All the possible outcomes for two dice are shown below:

Next section: Probability of A and B

Dice provide great illustrations for concepts in probability. The most commonly used dice are cubes with six sides. Here, we will see how to calculate probabilities for rolling three standard dice. It is a relatively standard problem to calculate the probability of the sum obtained by rolling two dice. There are a total of 36 different rolls with two dice, with any sum from 2 to 12 possible. How does the problem change if we add more dice?

Just as one die has six outcomes and two dice have 62 = 36 outcomes, the probability experiment of rolling three dice has 63 = 216 outcomes. This idea generalizes further for more dice. If we roll n dice then there are 6n outcomes.

We can also consider the possible sums from rolling several dice. The smallest possible sum occurs when all of the dice are the smallest, or one each. This gives a sum of three when we are rolling three dice. The greatest number on a die is six, which means that the greatest possible sum occurs when all three dice are sixes. The sum of this situation is 18.

When n dice are rolled, the least possible sum is n and the greatest possible sum is 6n.

• There is one possible way three dice can total 3
• 3 ways for 4
• 6 for 5
• 10 for 6
• 15 for 7
• 21 for 8
• 25 for 9
• 27 for 10
• 27 for 11
• 25 for 12
• 21 for 13
• 15 for 14
• 10 for 15
• 6 for 16
• 3 for 17
• 1 for 18

As discussed above, for three dice the possible sums include every number from three to 18. The probabilities can be calculated by using counting strategies and recognizing that we are looking for ways to partition a number into exactly three whole numbers. For example, the only way to obtain a sum of three is 3 = 1 + 1 + 1. Since each die is independent from the others, a sum such as four can be obtained in three different ways:

• 1 + 1 + 2
• 1 + 2 + 1
• 2 + 1 + 1

Further counting arguments can be used to find the number of ways of forming the other sums. The partitions for each sum follow:

• 3 = 1 + 1 + 1
• 4 = 1 + 1 + 2
• 5 = 1 + 1 + 3 = 2 + 2 + 1
• 6 = 1 + 1 + 4 = 1 + 2 + 3 = 2 + 2 + 2
• 7 = 1 + 1 + 5 = 2 + 2 + 3 = 3 + 3 + 1 = 1 + 2 + 4
• 8 = 1 + 1 + 6 = 2 + 3 + 3 = 4 + 3 + 1 = 1 + 2 + 5 = 2 + 2 + 4
• 9 = 6 + 2 + 1 = 4 + 3 + 2 = 3 + 3 + 3 = 2 + 2 + 5 = 1 + 3 + 5 = 1 + 4 + 4
• 10 = 6 + 3 + 1 = 6 + 2 + 2 = 5 + 3 + 2 = 4 + 4 + 2 = 4 + 3 + 3 = 1 + 4 + 5
• 11 = 6 + 4 + 1 = 1 + 5 + 5 = 5 + 4 + 2 = 3 + 3 + 5 = 4 + 3 + 4 = 6 + 3 + 2
• 12 = 6 + 5 + 1 = 4 + 3 + 5 = 4 + 4 + 4 = 5 + 2 + 5 = 6 + 4 + 2 = 6 + 3 + 3
• 13 = 6 + 6 + 1 = 5 + 4 + 4 = 3 + 4 + 6 = 6 + 5 + 2 = 5 + 5 + 3
• 14 = 6 + 6 + 2 = 5 + 5 + 4 = 4 + 4 + 6 = 6 + 5 + 3
• 15 = 6 + 6 + 3 = 6 + 5 + 4 = 5 + 5 + 5
• 16 = 6 + 6 + 4 = 5 + 5 + 6
• 17 = 6 + 6 + 5
• 18 = 6 + 6 + 6

When three different numbers form the partition, such as 7 = 1 + 2 + 4, there are 3! (3x2x1) different ways of permuting these numbers. So this would count toward three outcomes in the sample space. When two different numbers form the partition, then there are three different ways of permuting these numbers.

We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are:

• Probability of a sum of 3: 1/216 = 0.5%
• Probability of a sum of 4: 3/216 = 1.4%
• Probability of a sum of 5: 6/216 = 2.8%
• Probability of a sum of 6: 10/216 = 4.6%
• Probability of a sum of 7: 15/216 = 7.0%
• Probability of a sum of 8: 21/216 = 9.7%
• Probability of a sum of 9: 25/216 = 11.6%
• Probability of a sum of 10: 27/216 = 12.5%
• Probability of a sum of 11: 27/216 = 12.5%
• Probability of a sum of 12: 25/216 = 11.6%
• Probability of a sum of 13: 21/216 = 9.7%
• Probability of a sum of 14: 15/216 = 7.0%
• Probability of a sum of 15: 10/216 = 4.6%
• Probability of a sum of 16: 6/216 = 2.8%
• Probability of a sum of 17: 3/216 = 1.4%
• Probability of a sum of 18: 1/216 = 0.5%

As can be seen, the extreme values of 3 and 18 are least probable. The sums that are exactly in the middle are the most probable. This corresponds to what was observed when two dice were rolled.

1. Ramsey, Tom. “Rolling Two Dice.” University of Hawaiʻi at Mānoa, Department of Mathematics.

Answer

Hint: We will use the definition of the probability to solve this question. We will list all the possible outcomes and then we will use the formula of the probability as: Probability of an event = the total number of possible outcomes/ the total number of outcomes.

Complete step-by-step answer:

Hence, the probability of getting the total of two dice greater than 9 is $\dfrac{1}{3}$.

Note: You can also solve this question by using the method of conditional probability. It is better if you write all the outcomes and then select the required outcomes. Else, it will become a bit more confusing. And, always check that the probability obtained must lie between 0 and 1. Here as well, the probability is 0.333 which is between 0 and 1.

Both of them can be $5$ as well. So if the event of at least one $5$ is $B$ then,

$ \displaystyle P(B) = 2 \cdot \frac{1}{6} \cdot \frac{5}{6} + \frac 16 \cdot \frac 16 = \frac{11}{36}$

Or using complimentary method, $ \displaystyle P(B) = 1 - \frac 56 \cdot \frac 56 = \frac{11}{36}$

Now if $A$ is the event of sum being $10$ or more,

$ \displaystyle P(A \cap B) = 2 \cdot \frac 16 \cdot \frac 16 + \frac 16 \cdot \frac 16 = \frac{3}{36}$

As there are $3$ favorable outcomes out of $36$: $\{5, 5\}, \{5, 6\}, \{6, 5\}$

So, $ \displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} ~ $ does give you $\dfrac{3}{11}$