what is the chance of throwing at least 7 in a single cast with two dice
Answers
Answer:
21/36 = 0.58
Step-by-step explanation:
Assuming the two dice are unbiased, the total number of outcomes is 6^2 = 36
when a single die is thrown the number of outcomes is 6^1
when two dice are thrown the number of outcomes is 6^2
In general, when n number of dice are thrown the number of outcomes is 6^n
{
(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)
}
Every cell containing a number in bold, satisfies: (D1+D2)≥7 (Cause consider at least 7)
Here D1 = Outcome from first die
D2= Outcome from second die
Probability being defined as
(number of favorable outcome)/(number of total outcomes)
In this case chance or probability is = 21/36 = 0.58