Question

What is the probability that in three consecutive rolls of two fair dice, a person gets a total of 7, followed by a total of 11, followed by a total of 7?

Answer 1

Answer:

Step-by-step explanation:

Let us consider the first case. We denote the probability of "getting a total of 7 in a throw of 2 fair dices" by

P(A).

Now, there are 6 outcomes when we throw a dice, so the number of possible outcomes when we throw 2 dices at a time is

6⋅6=36

Now by observation, we get

1+6=7,2+5=7,3+4=7

They can interchange their position in

2! ways, so the number of all possible cases in favour of

A is

3⋅2=6

Hence

P(A)=636=16

Let us consider the 2nd case. We denote the probability of "getting a total of 11 in a throw of 2 fair dices" by

P(B).

Now, there are 6 outcomes when we throw a dice, so the number of possible outcomes when we throw 2 dices at a time is

6⋅6=62=36

Now by observation, we get

5+6=11

They can interchange their position in

2!ways, so the number of all possible cases in favour of

B is

1⋅2=2

Hence

P(B)=236=118

Let us consider the 3rd case. It is the same as the first case, hence

P(C)=636=16

The required probability asked in the question is

P

(

A

∩

B

∩

C

)

.

The events A, B, C are independent, so

P(A∩B∩C)=P(A)⋅P(B)⋅P(C)

=16⋅118⋅16=1/648