Answer:
The probability of getting A second number that is less than the first number is
[tex]P(E) = \frac{15}{36} = 0.4166[/tex]
Step-by-step explanation:
Step(i):-
Given a single die is rolled twice
In throwing a die , there are six exhaustive elementary events
1 or 2 or 3 or 4 or 5 or 6.
The total number of exhaustive events = 6
Given data a single die is rolled two times = 6² = 36
The total number of exhaustive cases n(S) = 36
Step(ii):-
Let 'E' be the event of getting A second number that is less than the first number.
The required pairs are
{(6,1),(6,2),(6,3),(6,4),(6,5),(5,4),(5,3),(5,2),(5,1),(4,3)(4,2),(4,1),(3,2),(3,1),(2,1)}
The total number of favorable cases n(E) = 15
The required probability
[tex]P(E) = \frac{n(E)}{n(S)} = \frac{15}{36}[/tex]
Conclusion:-
The probability of getting A second number that is less than the first number is
[tex]P(E) = \frac{15}{36} = 0.4166[/tex]
