in a simultaneous throw of pair of dice find the probability of getting
5 as the sum
2 will come up at least once
2 will not come either time
Answers
Step-by-step explanation:
Solution :-
Given that
A pair of dice thrown once
We know that
If n dice thrown simultaneously then the possible outcomes = 6^n
Number of dice = 2
So, The total number of possible outcomes
= 6^2 = 6×6 = 36
They are :
(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)
We know that
Probability of an event P (E) = Number of favourable outcomes/Total number of all possible outcomes
I) Probability of getting the sum as 5:-
Favourable outcomes to the sum as 5:
(1,4),(2,3),(3,2),(4,1)
Total number of favourable outcomes = 4
Probability of getting 5 as the sum on dice
= 4/36
= 1/9
ii) Probability of getting 2 will come up at least once :-
Favourable outcomes to 2 will come up at least once :
(1,2),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2)
Total number of favourable outcomes =11
Probability of getting 2 will come up at least once :
= 11/36
iii)Probability of getting 2 will not come
either time :-
Favourable outcomes to 2 will not come either time :
(1,1) ,(1,3),(1,4),(1,5),(1,6)
(3,1),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,3),(6,4),(6,5),(6,6)
Total number of favourable outcomes =25
Probability of getting 2 will not come either time
= 25/36
(or)
Let the probability of getting 2 will come once be P(E) then
Probability of not getting 2 will come either time=
P( not E)
We know that
P(E)+P( not E) = 1
=> P( not E) = 1-P(E)
=> P(not E) = 1-(11/36)
=> P(not E) = (36-11)/36
=> P( not E) = 25/36
Used formulae:-
- Probability of an event P (E) = Number of favourable outcomes/Total number of all possible outcomes
- If n dice thrown simultaneously then the possible outcomes = 6^n
- P(E)+P( not E) = 1