What is the probability that in three consecutive rolls of

Craig French

Craig French

Answered question

2022-02-11

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 & Explanation

illustrationsamk

illustrationsamk

Beginner2022-02-12Added 14 answers

Step 1
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=62=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 byobservation, 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(ABC)
The events A, B, C are independent, so
P(ABC)=P(A)×P(B)×P(C)=16×118×16=1648
narodiloxe4

narodiloxe4

Beginner2022-02-13Added 19 answers

Step 1
A possibility space is a good way of showing the possible outcomes when two dice are rolled:
1234561:2345672:3456783:4567894:56789105:678910116:7891011121234561:2345672:3456783:4567894:56789105:678910116:789101112
There are 6×6=36 possible outcomes
There are 6 ways of rolling 7 and 2 ways of rolling 11
P(7, 11, 7)=636×236×636
=16×118×16
=1648

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