Solving Dice Probability
Introduction to solving dice probability:
Let us see ,solving the dice probability. Probability is nothing but calculating the chance for a particular event to occur. Dice based problems are the best example for explaining about the probability.
Let E be an experiment involving rolling two dice and recording the value on top of each die. The notation for this sample space
is: S = {(i, j),i =1,2,3,4,5,6, j =1,2,3,4,5,6}
Note this is a discrete and finite sample space. Let us see the probability concepts, solving problems using the dice problems.
Solving Dice Probability:
Let us see some of the examples of solving dice probaility.
Example 1:
What is the probability of a die showing a 2 or a 5?
Solution:
P (2) = 1/6
P (5) = 1/6
P (2 or 5) = P(2) + P(5)
= (1/6) + (1/6)
= 2/6
= 1/3
The Probability of a die showing 2 or 5 is 1/3
Example 2:
In rolling two balanced dice, if the sum of the two values is 7, what is the probability that one of the values is 1?
Solution:
Event A is value of 1
Event B is sum equals 7
N AB = 2, {(1,6),(6,1)}
NB = 6
P(AB) = '2/36'
P(B) = 6/36
P(A | B) = P(AB) / P(B) = (2/36) / (6/36) = 1/3
Solving Dice Probability:
Example 3:
Three dice are rolled once. In this problem find the dice probability that the sum of the numbers on the two dice is greater than 10?
Solution:
When three dice are rolled, the sample space S = {(1, 1), (1, 2), (1, 3) ... (6, 6)}.
S contains 6 6 = 36 outcomes.
Let A be the event of probability of the sum of face numbers greater than or equal to 10.
A = {(6,6), (5,6), (6,5), (5,5)}.
n(Sample space S) = ' 216' , 'n(A) =' '4' .
Now let us use the calculate probability using probability formula,
P (A) = n(A)/n(S) = '4/216' ='1/54'
Example 4:
When 2 dice is thrown simultaneously once. Find the probability of getting a number lying between 5 and 11.
Solution:
Total number in possible outcomes associated with random experiment of throwing two dice is 12 ( that is 1, 2, 3, 4, 5, 6,7,8,9,10,11,12).
Let E be the event getting a number lying between 5 and 11.
Favorable number of elementary events (outcomes) = 5(i.e., 6, 7, 8, 9, 10)
P (E) = 5 / 12
by: nayaknandan
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