Science > Mathematics > Statistics and Probability > Probability > Problems Based on Throwing of Two Dice
In the last article, we have studied to solve problems to calculate probability when a single die is thrown. In this article, we are going to study to solve problems to find the probability involving the throwing of two dice.
Algorithm:
- Study experiment and write the sample space
- Find favourable point and write event space
- Use the definition of probability and find it.
Throwing of Two Dice:
Two fair dice are thrown Or a die is thrown twice
S = {1,2,3,4,5,6} × {1,2,3,4,5,6}
| 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 sum of the two numbers on two dice is called the score on two dice.
- The minimum score on two dice is 2 and the maximum score on two dice is 12.
- The cases favourable to a particular score can be read along the diagonal of that score.
Example – 01:
Two fair dice are tossed. Find the probability in the following cases:
Solution:
Two fair dice are thrown
The sample space is
S = {1,2,3,4,5,6} × {1,2,3,4,5,6}
| 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) | } |
∴ n (S) = 36
a) the sum of the scores is even
Let A be the event of that the sum of the scores is even i.e. 2, 4, 6, 8, 10, 12
∴ A = { (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1),
(3, 3), (3, 5), (4, 2), (4, 4), (4, 6),
(5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6) }
∴ n(A) = 18
By the definition P(A) = n(A)/n(S) = 18/36 = 1/2
Ans: the probability that the sum of the scores is even is 1/2
b) the sum of the scores is odd
Let B be the event of that the sum of the scores is odd i.e. 3, 5, 7, 9, 11
∴ B = { (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6),
(2, 5), (3, 4), (4, 3), (5, 2), (6, 1),
(3, 6), (4, 5), (5, 4), (6, 3), (6, 5), (5, 6) }
∴ n(A) = 18
By the definition P(B) = n(B)/n(S) = 18/36 = 1/2
Ans: the probability that the sum of the scores is odd is 1/2
c) the sum of the scores is a perfect square
Let C be the event of that the sum of the scores is a perfect square i.e. 4, 9.
∴ C = { (1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3) }
∴ n(C) = 7
By the definition P(C) = n(C)/n(S) = 7/36
Ans: the probability that the sum of the scores is a perfect square is 7/36
d) the sum of the score is a multiple of four or the score is divisible by 4
Let D be the event of that the sum of the score is a multiple of four i.e. 4, 8, 12
∴ D = { (1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6) }
∴ n(D) = 9
By the definition P(D) = n(D)/n(S) = 9/36 = 1/4
Ans: the probability that the sum of the score is a multiple of four is 1/4
e) the sum of the scores is a multiple of 3
Let E be the event of that the sum of the scores is a multiple of 3 i.e. 3, 6, 9, 12
∴ E = { (1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2),
(5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
∴ n(E) = 12
By the definition P(E) = n(E)/n(S) = 12/36 = 1/3
Ans: the probability that the sum of the score is a multiple of 3 is 1/3..
f) the sum of the points obtained is greater than 4
Let F be the event of that the sum of the points obtained is
greater than 4 i.e. 5, 6, 7, 8, 9, 10, 11, 12.
∴ F = { (1, 4), (1, 5), (1, 6) (2. 3), (2, 4), (2, 5). (2, 6), (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, 1I), (6, 2), (6, 3). (6, 4), (6, 5), (6, 6) }
∴ n(F) = 30
By the definition P(F) = n(F)/n(S) = 30/36 = 5/6
Ans: the probability that the sum of the points obtained is greater than 4 is 5/6
g) the sum of the points is at least 11
Let G be the event of that the sum of the points is at least 11 i.e. 11, 12
∴ G ={ (5, 6), (6, 5), (6, 6) }
∴ n(G) = 3
By the definition P(G) = n(G)/n(S) = 3/36 = 1/12
Ans: the probability that the sum of the points is at least 11 is 1/12.
h) the same score on the first die and second die
Let H be the event of that the same score on the first die and second die.
∴ H = { (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) }
∴ n(H) = 6
By the definition P(H) = n(H)/n(S) = 6/36 = 1/6
Ans: the probability that the same score on the first die and second die is 1/6.
i) the score on the second die is greater than the score on the first die
Let J be the event of that the score on the second die is greater than the score on the first die
∴ J = { (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2. 3), (2, 4), (2, 5).
(2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) }
∴ n(J) = 15
By the definition P(J) = n(J)/n(S) = 15/36 = 5/12
Ans: the probability that the score on the second die is greater than the score on the first die is 5/12.
j) the sum of the numbers on their faces obtained is either a perfect square or their sum is less than 5.
Let K be the event of that the sum of the numbers on their faces
obtained is either a perfect square or their sum is less than 5.
∴ K = { (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3. 1), (3, 6), (4, 5), (5, 4), (6, 3) }
∴ n(K) = 10
By the definition P(K) = n(K)/n(S) = 10/36 = 5/18
Ans: the probability that the sum of the numbers on their faces
obtained is either a perfect square or their sum is less than 5 is 5/18.
l) the sum of numbers shown is 7 or product is 12
Let M be the event of that the sum of numbers shown is 7 or product is 12
∴ M = { (1, 6), (2, 5), (2, 6), (3, 4), (4, 3), (5, 2), (6, 1), (6, 2) }
∴ n(M) = 8
By the definition P(M) = n(M)/n(S) = 8/36 = 2/9
Ans: the probability that the sum of numbers shown is 7 or the product is 12 is 2/9.
m) the sum of these scores is either a perfect square or a prime number
Let N be the event of that the sum of these scores is either
a perfect square or a prime number i.e. 4, 9, 2, 3, 5, 7, 11
∴ N = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 1), (1, 2),
(2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5),
(3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5) }
∴ n(N) = 22
By the definition P(N) = n(N)/n(S) = 22/36 = 11/18
Ans: the probability that the sum of numbers a perfect square or a prime number is 11/18.
n) the product of the scores is 12
Let Q be the event of that the product of the scores is 12
∴ Q = { (2, 6), (3, 4), (4, 3), (6, 2) }
∴ n(Q) = 4
By the definition P(Q) = n(Q)/n(S) = 4/36 = 1/9
Ans: the probability that the product of the scores is 12 is 1/9
o) the sum of these scores is either a perfect square or an even number
Let R be the event of that the sum of these scores is either
a perfect square or an even number i.e. 4, 9, 2, 6, 8,10, 12
∴ R = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 1), (1, 5),
(2, 4), (3, 3), (4, 2), (5, 1), (2, 6), (3, 5), (4, 4),
(5, 3), (6, 2), (4, 6), (5, 5), (6, 4), (6, 6) }
∴ n(R) = 22
By the definition P(R) = n(R)/n(S) = 22/36 = 11/18
Ans: the probability that the sum of these scores is either a perfect square or an even number is 11/18.
p) the sum of these scores is either an even number or a number divisible by 5
Let T be the event of that the sum of these scores is either
an even number or a number divisible by 5 i.e. 2, 4, 6, 8,10, 12, 5
∴ T = {(1. 1), (1, 3), (2, 2), (3, 1), (1, 5), (2, 4), (3, 3),
(4, 2), (5, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2),
(4, 6), (5, 5), (6, 4), (6, 6), (1, 4), (2, 3), (3, 2), (4, 1) }
∴ n(T) = 22
By the definition P(T) = n(T)/n(S) = 22/36 = 11/18
Ans: the probability that the sum of these scores is either
an even number or a number divisible by 5 is 11/18.
q) the sum of these scores is either a perfect square or a multiple of 3
Let T be the event of that the sum of these scores is either
a perfect square or a multiple of 3 i.e. 4, 9, 3, 6, 12
∴ R = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 2), (2, 1),
(2, 4), (3, 3), (4, 2), (5, 1), (6, 6) }
∴ n(T) = 14
By the definition P(T) = n(T)/n(S) = 14/36 = 7/18
Ans: the probability that the sum of these scores is either a perfect square or a multiple of 3 is 7/18.
r) the sum of the scores is either greater than 9 and an even number.
Let U be the event of that the sum of these scores is either
greater than 9 i.e. 10, 12
∴ U = {(4, 6), (5, 5), (6, 4), (6, 6) }
∴ n(U) = 4
By the definition P(U) = n(U)/n(S) = 4/36 = 1/9
Ans: the probability that the sum of these scores is either greater than 9 and an even number is 1/4.
s) the product of the scores is a perfect square
Let V be the event of that the product of the scores is
a perfect square. i.e. 1, 4, 9, 16, 25, 36
∴ V = {(1, 1), (1, 4), (2, 2), (4, 1), (3, 3), (4, 4), (5, 5), (6, 6) }
∴ n(V) = 8
By the definition P(V) = n(V)/n(S) = 8/36 = 2/9
Ans: the probability that the product of the scores is a perfect square. is 2/9.
In the next article, we shall study some basic problems of probability based on drawing a single card from the collection of numbered cards.