What is the probability that this sum is at least 5?

4 Answers By Expert Tutors

the sum of the dice is at least 5

one die is showing a 2

2 and 3

2 and 4

2 and 5

2 and 6

4 out of 36 ways to get the sum at least 5 and one die is always 2

Probability = 4/36 = 1/9

Hi,

there are 36 possible combinations rolling two die.

when I did the combinations given that one die is 2. I got 3,4,5,6 as possibilities for the second die. so, I got 4/36 or 1/9.

Hi Areli. Since one die already shows 2, you only have to focus on the other die.

That means to get at least 5 for the sum, the other die has to be 3, 4, 5, or 6. (4 possibilities).

For any die to begin with, the possibilities are 1 through 6 when you roll it. ( 6 possibilities). So you need 3, 4, 5, or 6, out of anywhere from 1 through 6. It means 4 possibilities out of 6, or 4/6. If you need to simplify, it becomes 2/3.

So the probability that the sum is at least 5 is 2/3.

Please consider hiring me as a tutor. Once my schedule is full, I won’t have as much time to provide this help. Thanks!

Hi Areli. Since one die already shows 2, you only have to focus on the other die.

That means to get at least 5 for the sum, the other die has to be 3, 4, 5, or 6. (4 possibilities).

For any die to begin with, the possibilities are 1 through 6 when you roll it. ( 6 possibilities). So you need 3, 4, 5, or 6, out of anywhere from 1 through 6. It means 4 possibilities out of 6, or 4/6. If you need to simplify, it becomes 2/3.

So the probability that the sum is at least 5 is 2/3.

Please consider hiring me as a tutor. Once my schedule is full, I won’t have as much time to provide this help. Thanks!

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Here is the sample space. It contains 36 possible outcomes:
 
(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)
 


Now I will color the ones red which have a sum of at least 5:
 


(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)


Count the red ones.  There are 30 of them. (it's easier to count the
6 non-red ones and subtract from 36 to get 30).

So the probability of a sum of at least 5 is 30 out of 36, which gives
us the fraction  which reduces to .

Edwin

Neither method is correct.

• In the first method, your outcomes are not equally likely

• In the second method, you're counting some outcomes twice.

In order to use the formula $$ \frac{\text{favorable outcomes}}{\text{total outcomes}}, $$ you need your events to be equally likely. In the first method, the outcomes $(1,1)$ and $(1,2)$ are not equally likely. The first method requires both dice to have a value $1$ while the second method has two situations for the dice.

To make this clearer, suppose that the dice are red and blue. Then, $(1,1)$ means that both the red die and the blue die show $1$. On the other hand, in the first method, $(1,2)$ represents the two possibilities ($1$-Red and $2$-Blue) or ($2$-Red and $1$-Blue). Since there are two possible ways to get a $1$ and a $2$, this $(1,2)$ has double the chances of occurring when compared to $(1,1)$.

For the second formulation, you're double counting the pairs of the form $(1,1)$. In this case, you're trying to describe $(1,1)$ for $1$-Red and $1$-Blue as well as $(1,1)$ for $1$-Blue and $1$-Red, but these are exactly the same situation.

Therefore, in the second case, you shouldn't duplicate the pairs that are identical under reversing the coordinates.

To calculate the probability correctly, the list should be $$ (1,1),(1,2),(1,3),(2,1),(2,2),(3,1). $$ Or, in other words, for the red and blue dice, $$ (1R,1B),(1R,2B),(1R,3B),(2R,1B),(2R,2B),(3R,1B). $$ Since there are $6$ possibilities for the red die and $6$ possibilities for the blue die, this results in $36$ total possible outcomes. Putting this all together, the probability is $6/36=1/6$.

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