How to solve a 3 dice problem?

Probability is also known as a possibility. This means math of chance, that trade in the happening of a likely event. The value is designated from zero to one. In math, Probability has been shown to guess how likely events are to occur. Basically, the probability is the scope to which something is to be anticipated to happen.

Probability

To understand probability more correctly, take an example as rolling a dice, the possible outcomes are 1, 2, 3, 4, 5, and 6. The possibility of occurring any of the favorable outcomes is 1/6. As the probability of occurring any of the events is the same so there are similar chances of getting any likely number, in this case, it is either 1/6 or 50/3.

Formula of Probability

Probability of an event = {Number of favourable events } ⁄ {number of total events}

P(A) = {Number of ways A occurs} ⁄ {Total number of events}

DICE 

Dice is a small block that has between one and six marks or dots on its edge and is used in games to give a random figure. Dice are small, tossable blocks with marked sides that can pause in several figures. They are used to give rise to random figures, often as part of sideboard games, as well as dice games, board games, role-playing games, and games of chance.

A customary die is a cube with each of its six faces noticeable with a different number of figures from one to six. When tossable or rolled, the die comes to pause appear a random number from one to six on its higher side, with the occurrence of each event being equally likely. Dice may also have concave or uneven shapes and may have faces marked with figures or characters instead of pips. Loaded dice are drawn to service some results over others for escape or entertainment.

How to solve a 3 dice problem?

Solution:

Possibility for throwing six sided three dice will be 1, 2, 3, 4, 5 and 6 dots in each (three) dies.

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Now, let’s consider the possible sums from rolling three dice. The smallest attainable sum occur when all of the dice are the smallest, or one each. This gives a sum of three when we are throwing three dice i.e., (1, 1, 1). The greatest number on a die is six, which means that the greatest possible sum happens when all three dice are sixes is 18 i.e., (6, 6, 6). When n dice are rolled, the minimal attainable sum is n and the greatest attainable sum is 6n. There is only one way when 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

Forming Sums

As shown above, for three dice the possible sums include every number from three to 18. Calculate probability by using add up plan and acknowledging that we are considering ways to separate a integer into absolutely three whole numbers. For example, to obtain a sum of three there is only one way i.e., 3 = 1 + 1 + 1. Since each die is individualistic from the others, a sum such as four can be acquired in three different ways:

• 1 + 1 + 2
• 1 + 2 + 1
• 2 + 1 + 1

Forward add up arguments can be used to find the number of ways of creating 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

Specific Probabilities

Probability of an event = {Number of favourable events } ⁄ {number of total events},or 216. The results are:

• Possibility of getting a sum of 3: 1/216 = 0.0046 × 100= 0.5%
• Possibility of getting a sum of 4: 3/216 = 0.0138 × 100 = 1.4%
• Possibility of getting a sum of 5: 6/216 = 0.0277 × 100 = 2.8%
• Possibility of getting a sum of 6: 10/216 = 0.0462 × 100 = 4.6%
• Possibility of getting a sum of 7: 15/216 = 0.069 × 100 = 7.0%
• Possibility of getting a sum of 8: 21/216 = 0.097 × 100 = 9.7%
• Possibility of getting a sum of 9: 25/216 = 0.115 × 100 = 11.6%
• Possibility of getting a sum of 10: 27/216 = 0.125 × 100 = 12.5%
• Possibility of getting a sum of 11: 27/216 = 0.125 × 100 = 12.5%
• Possibility of getting a sum of 12: 25/216 = 0.115 × 100 = 11.6%
• Possibility of getting a sum of 13: 21/216 = 0.097 × 100 = 9.7%
• Possibility of getting a sum of 14: 15/216 =0.069 × 100 = 7.0%
• Possibility of getting a sum of 15: 10/216 = 0.0462 × 100 = 4.6%
• Possibility of getting a sum of 16: 6/216 = 0.0277 × 100 = 2.8%
• Possibility of getting a sum of 17: 3/216 = 0.013 × 100 = 1.4%
• Possibility of getting a sum of 18: 1/216 = 0.0046 × 100 = 0.5%

As can be seen, the extreme values of 3 and 18 are least probable. The sums which are in the middle are most probable.

Sample Problems

Question 1: Three dice are thrown together. Find the probability of getting a total of 5.

Solution:

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of favourable events of getting a total of 5 = 6

i.e., (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2)

So, probability of occurring a total of 5

P(A) = {Number of favourable events } ⁄ {number of total events}

= 6/216

= 1/36

Question 2: Three dice are thrown together. Find the probability of getting a total of almost 5.

Solution:

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of affair of occurring a total of atmost 5 = 10

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2).

Therefore, possibility of happening a total of atmost 5

P(E) = {Number of favourable events } ⁄ {number of total events}  

= 10/216

= 5/108

Question 3: Three dice are thrown together. Find the probability of getting a total of at least 5.

Solution:

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of affair of happening a total of less than 5 = 4

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1).

Therefore, possibility of happening a total of less than 5

P(E) = {Number of favourable events } ⁄ {number of total events}  

= 4/216

= 1/54

Therefore, possibility of occurring a total of at least 5 = 1 – P (occurring a total of less than 5)

= 1 – 1/54

= (54 – 1)/54

= 53/54

Question 4: Three dice are thrown together. Find the probability of getting a total of 6.

Solution:

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of affair of happening a total of 6 = 10

i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).

Therefore, possibility of happening a total of 6

P(E) = {Number of favourable events } ⁄ {number of total events}

= 10/216

= 5/108

Question 5: Three dice are thrown together. Find the probability of getting a total of 8.

Solution:

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of affair of happening a total of 8 = 21

i.e. (2, 2, 4), (4, 2, 2), (2, 4, 2), (1, 5, 2), (2, 5, 1), (5, 1, 2), (2, 1, 5), (1, 2, 5), (3, 3, 2), (3, 2, 3), (2, 3, 3) and so on…

Therefore, possibility of happening a total of 8

P(E) = {Number of favourable events } ⁄ {number of total events}

= 21/216