The faces of a die bear numbers 0, 1, 2, 3, 4, 5. If the die is rolled twice, then find
the probability that the product of digits on the upper face is zero.
Answers
Step-by-step explanation:
Given:-
The faces of a die bear numbers 0, 1, 2, 3, 4, 5 and the die is rolled twice.
To find :-
Find the probability that the product of digits on the upper face is zero ?
Solution:-
The digits in the given die = 0,1,2,3,4,5
Number of digits = 6
We know that
A die is rolled n times or "n" dice rolled simultaneously then the total number of possible outcomes = 6^n
The die is rolled twice ,Then the number of all possible outcomes = 6^2=6×6 = 36
They are:
(0,0),(0,1),(0,2),(0,3),(0,4),(0,5)
(1,0), (1,1), (1,2), (1,3) ,(1,4), (1,5)
(2,0),(2,1),(2,2),(2,3),(2,4),(2,5)
(3,0),(3,1),(3,2),(3,3),(3,4),(3,5)
(4,0),(4,1),(4,2),(4,3),(4,4),(4,5)
(5,0),(5,1),(5,2),(5,3),(5,4),(5,5)
The favourable outcomes for getting zero on top face of the die=(0,0),(0,1),(0,2),(0,3),(0,4),(0,5),(1,0),(2,0),(3,0),(4,0),(5,0)
Total number of favourable outcomes = 11
We know that
The probability of an event E = P(E)
Number of favourable outcomes/ Total number of possible outcomes
Probability of getting zero on the top face of the die
= Number of favourable outcomes to zero/ Total number of possible outcomes
=> 11/36
Answer:-
Probability of getting zero on the top or upper face of the die = 11 / 36
Used formulae:-
- A die is rolled n times or "n" dice rolled simultaneously then the total number of possible outcomes = 6^n
- The probability of an event E = P(E)= Number of favourable outcomes / Total number of possible outcomes