Question

A pair of fair-dice is rolled. What is the probability that the second die lands on a higher value than the first?

Answer 1

Answer:

5/12

Step-by-step explanation:

A pair of fair-dice is rolled. What is the probability that the second die lands on a higher value than the first?

Total possible out comes are 36

Dice 1    Dice 2

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

Number of cases the second die lands on a higher value than the first

= 5 + 4 + 3 + 2 + 1 + 0

= 15

Probability that the second die lands on a higher value than the first = 15/36

= 5/12

Answer 2

Answer:

The required probability is

$\frac{5}{12}$

Step-by-step explanation:

Formula used:

Probability of an event E is

P(E)=(No.of favourable ways to E)/Total no. of ways

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

when a pair of fair-dice is rolled,

S={(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)}

n(S)=36

Let E be the event of higher value on the second die.

Then,

E={(1,2) (1,3), (1,4), (1,5), (1,6)

(2,3), (2,4), (2,5), (2,6)

(3,4), (3,5), (3,6)

(4,5), (4,6)

(5,6)}

n(E)=15

Now,

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

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

$P(E)=\frac{5}{12}$