Two dice are rolled. the probability that one or both numbers on the faces will be 4 is

Whether you’re wondering what your chances of success are in a game or are just preparing for an assignment or exam on probabilities, understanding dice probabilities is a good starting point. Not only does it introduce you to the basics of calculating probabilities, it’s also directly relevant to craps and board games. It's easy to figure out the probabilities for dice, and you can build your knowledge from the basics to complex calculations in just a few steps.

Probabilities are calculated using the simple formula:

Probability = Number of desired outcomes ÷ Number of possible outcomes

So to get a 6 when rolling a six-sided die, probability = 1 ÷ 6 = 0.167, or 16.7 percent chance.

Independent probabilities are calculated using:

Probability of both = Probability of outcome one × Probability of outcome two

So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0.0278, or 2.78 percent.

The simplest case when you're learning to calculate dice probability is the chance of getting a specific number with one die. The basic rule for probability is that you calculate it by looking at the number of possible outcomes in comparison to the outcome you’re interested in. So for a die, there are six faces, and for any roll, there are six possible outcomes. There is only one outcome you’re interested in, no matter which number you choose.

\text{Probability} = \frac{\text{Number of desired outcomes}}{\text{Number of possible outcomes}}

For the odds of rolling a specific number (6, for example) on a die, this gives:

\text{Probability} = 1 ÷ 6 = 0.167

Probabilities are given as numbers between 0 (no chance) and 1 (certainty), but you can multiply this by 100 to get a percentage. So the chance of rolling a 6 on a single die is 16.7 percent.

If you’re interested in rolls of two dice, the probabilities are still simple to work out. If you want to know the likelihood of getting two 6s when you roll two dice, you are calculating “independent probabilities.” This is because the result of one die doesn’t depend on the result of the other die at all. This essentially leaves you with two separate one-in-six chances.

The rule for independent probabilities is that you multiply the individual probabilities together to get your result. As a formula, this is:

\text{Probability of both} = \text{Probability of outcome one} × \text{Probability of outcome two}

This is easiest if you work in fractions. For rolling matching numbers (two 6s, for example) from two dice, you have two 1/6 chances. So the result is:

\text{Probability} = \frac{1}{6} × \frac{1}{6} = \frac{1}{36}

To get a numerical result, you complete the final division:

\frac{1}{36}=1 ÷ 36 = 0.0278

As a percentage, this is 2.78 percent.

This gets a bit more complicated if you’re looking for the probability of getting two specific different numbers on two dice. For example, if you’re looking for a 4 and a 5, it doesn’t matter which die you roll the 4 with or which you roll the 5 with. In this case, it’s best to just think about it as in the previous section. Out of the 36 possible results, you’re interested in two outcomes, so:

\text{Probability} = \frac{\text{Number of desired outcomes}}{\text{Number of possible outcomes}} = \frac{2}{36} = 0.0556

As a percentage, this is 5.56 percent. Note that this is twice as likely as rolling two 6s.

If you want to know how likely it is to get a certain total score from rolling two or more dice, it’s best to fall back on the simple rule: Probability = Number of desired outcomes ÷ Number of possible outcomes. As before, you determine the total outcome possibilities by multiplying the number of sides on one die by the number of sides on the other. Unfortunately, counting the number of outcomes you’re interested in means a little bit more work.

For getting a total score of 4 on two dice, this can be achieved by rolling a 1 and 3, 2 and 2, or a 3 and 1. You have to consider the dice separately, so even though the result is the same, a 1 on the first die and a 3 on the second die is a different outcome from a 3 on the first die and a 1 on the second die.

For rolling a 4, we know there are three ways to get the outcome desired. As before, there are 36 possible outcomes. So we can work this out as follows:

\text{Probability} = \frac{\text{Number of desired outcomes}}{\text{Number of possible outcomes}} = \frac{3}{36}=0.0833

As a percentage, this is 8.33 percent. For two dice, 7 is the most likely result, with six ways to achieve it. In this case, probability = 6 ÷ 36 = 0.167 = 16.7 percent.

$11$ out of $36$? I got this by writing down the number of possible outcomes ($36$) and then counting how many of the pairs had a $6$ in them: $(1,6)$, $(2,6)$, $(3,6)$, $(4,6)$, $(5,6)$, $(6,6)$, $(6,5)$, $(6,4)$, $(6,3)$, $(6,2)$, $(6,1)$. Is this correct?

Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die.

Probability – Sample space for two dice (outcomes):

Note: 

(i) The outcomes (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) are called doublets.

(ii) The pair (1, 2) and (2, 1) are different outcomes.

Worked-out problems involving probability for rolling two dice:

1. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of 4 respectively. Then, show that

(i) A is a simple event

(ii) B and C are compound events

(iii) A and B are mutually exclusive

Solution:

Clearly, we haveA = {(1, 1)}, B = {(1, 2), (2, 1)} and C = {(1, 3), (3, 1), (2, 2)}.

(i) Since A consists of a single sample point, it is a simple event.

(ii) Since both B and C contain more than one sample point, each one of them is a compound event.

(iii) Since A ∩ B = ∅, A and B are mutually exclusive.

2. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5, and B is the event that at least one of the dice shows up a 3. Are the two events (i) mutually exclusive, (ii) exhaustive? Give arguments in support of your answer.

Solution:

When two dice are rolled, we have n(S) = (6 × 6) = 36.

Now, A = {(1, 4), (2, 3), (4, 1), (3, 2)}, and

B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1,3), (2, 3), (4, 3), (5, 3), (6, 3)}

(i) A ∩ B = {(2, 3), (3, 2)} ≠ ∅.

Hence, A and B are not mutually exclusive.

(ii) Also, A ∪ B ≠ S.

Therefore, A and B are not exhaustive events.

More examples related to the questions on the probabilities for throwing two dice.

3. Two dice are thrown simultaneously. Find the probability of:

(i) getting six as a product

(ii) getting sum ≤ 3

(iii) getting sum ≤ 10

(iv) getting a doublet

(v) getting a sum of 8

(vi) getting sum divisible by 5

(vii) getting sum of atleast 11

(viii) getting a multiple of 3 as the sum

(ix) getting a total of atleast 10

(x) getting an even number as the sum

(xi) getting a prime number as the sum

(xii) getting a doublet of even numbers

(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die

Solution: 

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36.

(i) getting six as a product:

Therefore, probability of getting ‘six as a product’

(ii) getting sum ≤ 3:

Therefore, probability of getting ‘sum ≤ 3’

(iii) getting sum ≤ 10:

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

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

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

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

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

(6, 1), (6, 2), (6, 3), (6, 4)] = 33

Therefore, probability of getting ‘sum ≤ 10’

(iv) getting a doublet: Let E4 = event of getting a doublet. The number which doublet will be E4 = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6

Therefore, probability of getting ‘a doublet’

(v) getting a sum of 8:

Therefore, probability of getting ‘a sum of 8’

(vi) getting sum divisible by 5:

Therefore, probability of getting ‘sum divisible by 5’

(vii) getting sum of atleast 11:

Therefore, probability of getting ‘sum of atleast 11’

(viii) getting a multiple of 3 as the sum:

Therefore, probability of getting ‘a multiple of 3 as the sum’

(ix) getting a total of atleast 10:

Therefore, probability of getting ‘a total of atleast 10’

(x) getting an even number as the sum:

Therefore, probability of getting ‘an even number as the sum

(xi) getting a prime number as the sum:

Therefore, probability of getting ‘a prime number as the sum’

(xii) getting a doublet of even numbers:

Therefore, probability of getting ‘a doublet of even numbers’

(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die:

Therefore, probability of getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’

4. Two dice are thrown. Find (i) the odds in favour of getting the sum 5, and (ii) the odds against getting the sum 6.

Solution:

We know that in a single thrown of two die, the total number of possible outcomes is (6 × 6) = 36.

Let S be the sample space. Then, n(S) = 36.

(i) the odds in favour of getting the sum 5:

(ii) the odds against getting the sum 6:

5. Two dice, one blue and one orange, are rolled simultaneously. Find the probability of getting 

(i) equal numbers on both 

(ii) two numbers appearing on them whose sum is 9.

Solution:

The possible outcomes are 

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

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

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

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

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

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

Therefore, total number of possible outcomes = 36.

(i) Number of favourable outcomes for the event E

                   = number of outcomes having equal numbers on both dice 

                   = 6    [namely, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)].

So, by definition, P(E) = \(\frac{6}{36}\)

                                 = \(\frac{1}{6}\)

(ii) Number of favourable outcomes for the event F

           = Number of outcomes in which two numbers appearing on them have the sum 9

            = 4     [namely, (3, 6), (4, 5), (5, 4), (3, 6)].

Thus, by definition, P(F) = \(\frac{4}{36}\)

                                    = \(\frac{1}{9}\).

These examples will help us to solve different types of problems based on probability for rolling two dice.

• 

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  Probability in everyday life, we come across statements such as: Most probably it will rain today. Chances are high that the prices of petrol will go up. I doubt that he will win the race. The words ‘most probably’, ‘chances’, ‘doubt’ etc., show the probability of occurrence

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  Practice different types of rolling dice probability questions like probability of rolling a die, probability for rolling two dice simultaneously and probability for rolling three dice simultaneously in rolling dice probability worksheet. 1. A die is thrown 350 times and the

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