In this chapter, we will solve questions related to throwing of two dice.
We know that in a dice there are six numbers ( 1 to 6 ). When we throw two dice simultaneously, there are total of 6 x 6 = 36 possible outcomes.
The sample space for experiment is given as;
S = { (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 outcome of getting same number on both the dice is called doublets.
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
I hope you understood the basic concept. Let’s solve some problems;
Questions on two dice probability
Question (01)
Two dice are thrown simultaneously. Find the probability of getting (1, 1) and (1, 5).
Solution
Total possible outcome = 36
We want to get (1, 1) or (1, 5) as final solution.
Number of favorable outcome = 2
Probability = Number of favorable outcome / total possible outcome
Probability = 2 / 36 = 1/18
Hence, 1/18 is the required probability
Question 02
Two dice are thrown simultaneously. Find the probability of getting doublets.
Solution
When two dice are thrown, the doublets are;
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Number of favorable outcome = 6
Total possible outcomes = 36
Probability = 6 / 36 = 1 / 6
Hence, 1/6 is the required probability.
Question 03
In an experiment of throwing two dice simultaneously, find the probability for;
(a) getting sum of 4
(b) getting sum of 9
Solution
(a) Let A be the event to get sum of 4
The event A is represented as;
A = { (1, 3), (2, 2), (3, 1) }
Number of favorable outcome = 3
Total possible outcome = 36
Probability = 3 / 36 = 1/12
(b) Let B be the event of getting sum 9
B = { (3, 6), (4, 5), (5, 4), (6, 3) }
Number of favorable outcome = 4
Total number of possible outcome = 36
Probability = 4 / 36 = 1/9
Hence, 1/9 is the required probability.
Question 04
In an experiment of throwing two dice simultaneously, Find the probability of getting even number on first dice.
Solution
Let A be the event of getting even number on first dice.
A = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
Number of favorable outcome = 18
Total possible outcome = 36
Probability = 18 / 36 = 1/2
Hence, 1/2 is the probability of getting even number on first dice.
Question 05
Two dice are thrown simultaneously. Find the following probability;
(a) getting sum divisible by 7
(b) getting sum at least 11
(c) getting even multiple on first and odd multiple on second dice
Solution
(a) getting sum divisible by 7
Let A be the even of getting sum divisible by 7.
A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) }
Total possible outcome = 6
Number of favorable outcome = 36
Probability = 6 / 36 = 1/6
(b) getting sum at least 11
Let B be the event of getting sum atleast 11.
B = {(5, 6), (6, 5), (6 , 6) }
Number of favorable outcome = 3
Total possible outcome = 36
Probability = 3/36 = 1/12
(c) getting even multiple on first and odd multiple on second dice
C = {(2, 1), (2, 3), (2, 5),
(4, 1), (4, 3), (4, 5),
(6, 1), (6, 3), (6, 5),}
Number of favorable outcome = 9
Total possible outcome = 36
Probability = 9/36 = 1/4
Hence, 1/4 is the required probability.