Two fair dies are rolled. What is the conditional probability thatat least one lands on 6 given that the dies land on different numbers?

33.33% conditional probability thatat least one lands on 6 given that the dies land on different numbers

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

All outcoms of the dices:

Format(Dice A, Dice B)

(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)

36 in all

Total outcomes:

In this question, we want all with no repetition.

There are 6 repetitions, they are (1,1), (2,2), ..., (6,6). So 36 = 6 = 30 outcomes.

Desired outcomes:

One landing on six:

(6,1), (6,2), (6,3), (6,4), (6,5), (1,6), (2,6), (3,6), (4,6), (5,6).

10 desired outcomes.

Probability:

10/30 = 0.3333

33.33% conditional probability thatat least one lands on 6 given that the dies land on different numbers

step-by-step explanation: