Question

A single die is rolled twice. The 36​ equally-likely outcomes are shown to the right. Find the probability of getting A second number that is less than the first number

Answer 1

Answer:

The probability of getting A second number that is less than the first number is

$P(E) = \frac{15}{36} = 0.4166$

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

  $P(E) = \frac{n(E)}{n(S)} = \frac{15}{36}$

Conclusion:-

The probability of getting A second number that is less than the first number is

$P(E) = \frac{15}{36} = 0.4166$