Question

3. Rolling two fair dice.a.List the sample space.b. Let X,: faced value of the 1° die; X2: faced value of the 2nd die;Y: the sum of two dice. Find:(i) Pr (X1 ≥ 5 n X2 = 2)(ii) Pr (X1 = 3 U X2 = 3)(ini) Pr (X1 ≥ 3 | Y = 8)(iv) Pr (X2 < 5 | Y > 7)

Answer 1

Q.3: Rolling two fair dice

a. Sample space

Each die has 6 faces and we have 2 die

There are a total of 6*6 = 36 possible outcomes.

Sample space = { (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) }

Part b:

X₁ is the face value of the 1st die

X₂ is the face value of the 2nd die

Y is the sum of two dice

(i) Pr (X₁ ≥ 5 ∩ X₂ = 2)

The probability that X₁ is equal to or greater than 5 is given by

There are 2 outcomes of interest X₁ = 5 and X₁ = 6 and the total number of outcomes are 6.

`P(X_1\ge5)=(2)/(6)=(1)/(3)`

So, Pr (X₁ ≥ 5) = 1/3

The probability that X₂ is equal 2 is given by

There is only 1 outcome of interest that is X₂ = 2 and the total number of outcomes are 6.

`P(X_2=2)=(1)/(6)`

So, Pr (X₂ = 2) = 1/6

The intersection of these two probabilities is given by

`P(X_1\ge5\cap X_2=2)=(1)/(3)*(1)/(6)=(1)/(18)`

Therefore, Pr (X₁ ≥ 5 ∩ X₂ = 2) = 1/18

(ii) Pr (X₁ = 3 U X₂ = 3)

The probability that X₁ is equal 3 is given by

There is only 1 outcome of interest that is X₁ = 2 and the total number of outcomes are 6.

`P(X_1=3)=(1)/(6)`

So, Pr (X₁ = 3) = 1/6

The probability that X₂ is equal 3 is also the same.

So, Pr (X₂ = 3) = 1/6

The union of these two probabilities is given by

`P(X_1=3\cup X_2=3)=P(X_1=3)+P(X_2=3)-P(X_1=3\cap X_2=3)`

Where P(X₁ = 3 ∩ X₂ = 3) is the intersection of the two probabilities.

This is the event when there is a 3 on both of the dice.

P(X₁ = 3 ∩ X₂ = 3) = 1/36

`P(X_1=3\cup X_2=3)=(1)/(6)+(1)/(6)-(1)/(36)=(11)/(36)`

Therefore, Pr (X₁ = 3 U X₂ = 3) = 11/36

(iii) Pr (X₁ ≥ 3 | Y = 8)

This is a conditional probability which is given by

`P(X_1\ge3|Y=8)=(P(X_1\ge3\cap Y=8))/(P(Y=8))`

Let us find each of these probabilities

The probability that X₁ is equal to or greater than 3 is given by

There are 4 outcomes of interest X₁ = 3, X₁ = 4, X₁ = 5, and X₁ = 6 and the total number of outcomes are 6.

`P(X_1\ge3)=(4)/(6)=(2)/(3)`

So, Pr (X₁ ≥ 3) = 2/3

The probability that Y = 8 is given by

Count the number of outcomes that will give us Y = 8

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

So, there are 5 outcomes of interest and the total number of outcomes are 36.

`P(Y=8)=(5)/(36)`

So, Pr (Y = 8) = 5/36

The intersection of these two probabilities is given by

`P(X_1\ge3\cap Y=8)=(2)/(3)*(5)/(36)=(5)/(54)`

So, the conditional probability is

`P(X_1\ge3|Y=8)=(P(X_1\ge3\cap Y=8))/(P(Y=8))=((5)/(54))/((5)/(36))=(5)/(54)*(36)/(5)=(2)/(3)`

Therefore, Pr (X₁ ≥ 3 | Y = 8) = 2/3

(iv) Pr (X₂ < 5 | Y > 7)

This is a conditional probability which is given by

`P(X_2<5|Y>7)=(P(X_2<5\cap Y>7))/(P(Y>7))`

The probability that X₂ is less than 5 is given by

There are 4 outcomes of interest X₂ = 4, X₂ = 3, X₂ = 2, and X₂ = 1 and the total number of outcomes are 6.

`P(X_2<5)=(4)/(6)=(2)/(3)`

The probability that Y > 7 is given by

Count the number of outcomes that will give us Y > 7

So, there are 16 outcomes of interest and the total number of outcomes are 36.

`P(Y>7)=(16)/(36)=(4)/(9)`

So, Pr (Y > 7) = 4/9

The intersection of these two probabilities is given by

`P(X_2<5\cap Y>7)=(2)/(3)*(4)/(9)=(8)/(27)`

So, the conditional probability is

`P(X_2<5|Y>7)=(P(X_2<5\cap Y>7))/(P(Y>7))=((8)/(27))/((4)/(9))=(8)/(27)*(9)/(4)=(2)/(3)`

Therefore, Pr (X₂ < 5 | Y > 7) = 2/3