Total numbers of elementary events are: 6 x 6 = 36
Let E be the event of getting an even number on both the dice
Favorable outcomes are: [2,2] , [2,4], [2,6], [4,2], [4,4], [4,6], [6,2], [6,4], [6,6]
Numbers of favorable outcomes are = 9
P [getting even number on both dice] = P [E] = \[\frac{9}{36}\] = \[\frac{1}{4}\]
`1/6``1/2``3/4``5/6`
Solution : When two dice are thrown simultaneously, all 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],[4,5],[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].
Number of all possible outcomes = 36.
[i] Let `E_[1]` be the event of getting two numbers whose sum is 5.
Then, the favourable outcomes are [1,4],[2,3],[3,2],[4,1].
Number of favourable outcomes = 4.
`:.` P [getting two numbers whose sum is 5] = `P[E_[1]] = 4/36 = 1/9`.
[ii] Let `E_[2]` be the event of getting a even numbers on both dice.
Then, the favourable outcomes are
[2,2],[2,4],[2,6],[4,2],[4,4],[4,6],[6,2],[6,4],[6,6].
Number of favourable outcomes = 9.
`:. ` P[getting even number on both dice ] = `P[E_[2]] = 9/36 = 1/4`.
[iii] Let `E_[3]` be the event of getting a doublet.
Then, the favourable outcomes are
[1,2],[2,2],[3,3],[4,4],[5,5],[6,6].
Number of favourable outcomes = 6.
`:. ` P[getting a doublet ] = `P[E_[3]] = 6/36 = 1/6`.
The term probability refers to computing the chance that certain events will happen. The likelihood of dice being a specific digit is 1 / 6. The probability associated with one dice roll is given as follows.
One Dice Roll
The uncomplicated scenario of dice probability is the likelihood of obtaining a specific number with a single dice. In the probability, the fundamental rule is that an individual must compute it by seeing the count of feasible end results in contrast to the desired end result. A dice consists of 6 feasible outcomes. The interest of an individual would be for an end result regardless of the choice of the digit. The sample space in this case = S = {1, 2, 3, 4, 5, 6}.
The formula used in this case is
Probability = count of favourable end results / count of total possible outcomes
The odds of spinning a specific number, if the digit is 6, this provides:
Probability = 1 / 6 = 0.167
The concept of probability is accessible as numerals between no likelihood and sureness. No chance or likelihood refers to 0 and sureness refers to 1. A person can multiply it by the number 100 to arrive at the percentage. Subsequently, the likelihood of spinning the digit 6 on the dice is 16.7%.
Two or More Dice
The computation of probabilities is a little difficult when it involves 2 fair dice. The estimation of independent probabilities happens when an individual wants to find the probability of obtaining 6 twice on spinning two dice. The outcome of one dice is independent of the outcomes of the other dice. Independent probabilities can be found by finding the product of the individual probabilities. The formula is
Probability of the two together = Probability of end result 1 * Probability of end result 2
The sample space when two dice are rolled is given below.
[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]
36 feasible outcomes exists.
[i] The end results [1, 1], [2, 2], [3, 3], [4, 4], [5, 5] and [6, 6] are termed doublets.
[ii] The outcomes [1, 2] and [2, 1] are different end results.
The following data displays the probabilities when a certain number is rolled on two-dice.
2 – 1 / 36 [2.778%]
3 – 2 / 36 [5.556%]
4 – 3 / 36 [8.333%]
5 – 4 / 36 [11.111%]
6 – 5 / 36 [13.889%]
7 – 6 / 36 [16.667%]
8 – 5 / 36 [13.889%]
9 – 4 / 36 [11.111%]
10 – 3 / 36 [8.333%]
11 – 2 / 36 [5.556%]
12 – 1 / 36 [2.778%]
The likelihood of spinning a particular digit or less for two 6-faced dice is given by
2 – 1 / 36 [2.778%]
3 – 3 / 36 [8.333%]
4 – 6 / 36 [16.667%]
5 – 10 / 36 [27.778%]
6 – 15 / 36 [41.667%]
7 – 21 / 36 [58.333%]
8 – 26 / 36 [72.222%]
9 – 30 / 36 [83.333%]
10 – 33 / 36 [91.667%]
11 – 35 / 36 [97.222%]
12 – 36 / 36 [100%]
Total result from 2 or more dice
If a person needs to obtain the chance of obtaining a specific total score on spinning two or more dice, then the following rule must be followed.
Probability = count of favourable end results / count of total possible outcomes
If a person wants a total score of 4 on two dice, it can be accomplished by spinning 1 & 3, 3 & 1, or 2 & 2.
It is mandatory to take the dice individually. The number 1 on the first dice and 3 on the second dice are not the same as number 3 on the initial dice and 1 on the other dice.
For spinning the number 4, 3 methods are present.
Probability = count of favourable end results / count of total possible outcomes
= 3 / 36
= 0.0833.
= 8.33%
To calculate the probability of obtaining a total of number 7, there exist 6 ways to accomplish it.
In this case, the probability is calculated as follows.
Probability = count of favourable end results / count of total possible outcomes
= 6 / 36
= 0.167
= 16.7%
Solved Problems On Dice Probability Formula
Problem 1: A single dice is rolled, what is the probability of obtaining a number that is even or a digit less than 5?
Answer:
The sample space when a single dice is rolled = S = {1, 2, 3, 4, 5, 6}.
Probability = count of favourable end results / count of total possible outcomes
Let E be the event of obtaining an even number and F be the event of getting a number less than 5.
E = {2, 4, 6}
F = {1, 2, 3, 4}
To find E or F.
E ⋃ F = {1, 2, 3, 4, 6}
It consists of 5 elements.
P [E ⋃ F] = 5 / 6
Problem 2: 2 fair dice are rolled. Find the probabilities of obtaining an even and an odd number.
Answer:
The sample space when two dice are rolled is given as follows.
[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]
The outcomes containing an even and an odd number are E=\{[2,1],[4,1],[6,1],[1,2],[3,2],[5,2],[2,3],[4,3],\\ [6,3],[1,4],[3,4],[5,4],[2,5],[4,5],[6,5],[1,6],[3,6],[5,6]\}.\\ \text { There are 18 elements in E, the probability is given by}\\ P[\text { One Even and One Odd}]=\frac{18}{36}=\frac{1}{2}