Question

Two regular six-sided dice are tossed. Compute the probability that the sum of the pips on the upward faces of the two dice is the following. (See the figure below for the sample space of this experiment. Enter the probability as a fraction.) At most 5

Answer 1

Answer:

We can find the probability to obtain two number that sum 7, 6, 5, 4, 3, or 2.

To find this, we are gonna use the regular probability definition, which is the quotient between the possible events and total number of events.

So, to find the probabilty of getting to number that sum 7, we can analyse the situation by written down to rows that shows all possible number in each dice

Dice 1     Dice 2

    1              1

    2             2

    3             3

    4             4

    5             5

    6             6

So, you need to group each pair of number that sum 7, those would be

1 + 6 = 7

2 + 5 = 7

3 + 4 = 7

4 + 3 = 7

5 + 2 = 7

6 + 1 = 7

There are 6 possible events that sum seven, so the probability of getting seven is

$P_{7}= \frac{6}{36} = \frac{1}{6}$, where the total number events are 36, because there are 36 possible sums.

If you repeat this process to all cases, that is, to sum 6, 5, 4, 3, 2, you would find that there are 5 possible events to obtain a sum results of six, there are 4 possible eventes to obtain five, 3 events to obtain two and 1 event to obtain 2.

So, all those probabilities would be

$P_{6}=\frac{5}{36}\\ P_{5}=\frac{4}{36}=\frac{2}{18}=\frac{1}{9}\\ P_{4}=\frac{3}{36}=\frac{1}{12}\\ P_{3}=\frac{2}{36}=\frac{1}{18} \\P_{2}=\frac{1}{36}$

Answer 2

Answer:

1/9

Step-by-step explanation:

The first die has 6 possible outcomes.  For each outcome there are 6 possible outcomes for the second die.  So the total number of combinations is 6×6 = 36.

The possible combinations that are at most 5:

1 and 4, 2 and 3, 3 and 2, 4 and 1

So the probability is 4/36 = 1/9.