Question

Two dice are thrown 3 times. Determine the probability that the difference of the numbers on the two dice is 2

Answer 1

Step-by-step explanation:

Sample Space(total outcomes):36

Favourable outcomes:=A= 8{(1,3),(3,1),(2,4),(4,2),(3,5),(5,3),(4,6),(6,4)}

since P(E)=no. of favourable outcomes divided by sample Space

therefore P(A)=8/36=2/9

Answer 2

The probability that the difference of the numbers on the two dice is 2 will be $\dfrac{8}{729}$

Step-by-step explanation:

Sample space of two dice thrown: 36

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

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

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

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

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

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

Outcomes difference of the numbers on the two dice is 2:

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

Number of outcomes = 8

$\text{Probability(difference of number 2)}=\dfrac{\text{Outcomes}}{\text{Total sample space}}$

$\text{Probability(difference of number is 2)}=\dfrac{8}{36}$

Probability(difference on two dice is 2 three trial)=P(first trial)×P(Second trial)×P(Third trial)

$\text{Probability(difference of number is 2)}=\dfrac{8}{36}\times \dfrac{8}{36}\times \dfrac{8}{36}$

$\text{Probability(difference of number is 2)}=\dfrac{8}{729}$

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