Question #9140

Two dice are thrown. What is the conditional probability of a total score of at least 8, given that one of the dice has thrown 4 .

Expert's answer

Consider the following events:

X = The total score is 8

Y = one of

the dice has thrown 4.

We should find the probability P(X | Y).

By

definition

P(X | Y) = P(XY) / P(Y)

Let us compute P(Y).

Let

A be the value of the first dice, and B be the value of second dice.

Then

there are 36 possible pairs (A,B).

Among them there are 10 pairs

containing 4:

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

(4,1), (4,2),

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

Each pair has teh same probability, therefore

P(Y) = 10/36.

Now compute P(XY).

Notice that

XY = The

total score is 8 and one of the dice has thrown 4.

This implies that

another dice has thrown 8-4=4, and so XY corresponds to a unique pair

(4,4).

Hence

P(XY) = 1/36.

Thus we obtain that

P(X

| Y) = P(XY) / P(Y) = (1/36) / (10/36) = 1/10 = 0.1.

X = The total score is 8

Y = one of

the dice has thrown 4.

We should find the probability P(X | Y).

By

definition

P(X | Y) = P(XY) / P(Y)

Let us compute P(Y).

Let

A be the value of the first dice, and B be the value of second dice.

Then

there are 36 possible pairs (A,B).

Among them there are 10 pairs

containing 4:

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

(4,1), (4,2),

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

Each pair has teh same probability, therefore

P(Y) = 10/36.

Now compute P(XY).

Notice that

XY = The

total score is 8 and one of the dice has thrown 4.

This implies that

another dice has thrown 8-4=4, and so XY corresponds to a unique pair

(4,4).

Hence

P(XY) = 1/36.

Thus we obtain that

P(X

| Y) = P(XY) / P(Y) = (1/36) / (10/36) = 1/10 = 0.1.

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