Question

Franklin rolls a pair of six-sided fair dice with sides numbered 1 through 6. The probability that the sum of the numbers rolled is either even or a multiple of 5 is ___. The probability that the sum of the numbers rolled is either a multiple of 3 or 4 is ___.

Answer 1

Answer: The required probabilities are

(1) $\dfrac{11}{18}.$

(2) $\dfrac{5}{9}.$

Step-by-step explanation:  Given that Franklin rolls a pair of six-sided fair dice with sides numbered 1 through 6.

Let 'S' be the sample space for the experiment.

Then, S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2),  .  .  ., (6, 5), (6,6) }

That is, n(S) = 36.

(1) We are to find the probability that the sum of the numbers rolled is either even or a multiple of 5.

Let 'A' and 'B' be the events that the sum of the numbers rolled is an even number and a multiple of 5 respectively.

Then,

A = {(1,1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4,2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)},

and

B = {(1,4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}.

So, the event that the sum rolled is an even number or a multiple of 5 is given by

A ∪ B = {(1,1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4,2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6), (1,4), (2, 3), (3, 2), (4, 1)}

⇒ n(A ∪ B) = 22.

Therefore, the probability that the sum of the numbers rolled is either even or a multiple of 5 will be

$P(A\cup B)=\dfrac{n(A\cup B)}{n(S)}=\dfrac{22}{36}=\dfrac{11}{18}.$

(2) We are to find the probability that the sum of the numbers rolled is either a multiple of 3 or a multiple of 4.

Let 'C' and 'D' be the events that the sum of the numbers rolled is a multiple of 3 and a multiple of 4 respectively.

Then,

C = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)},

and

D = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}.

So, the event that the sum rolled is a multiple of 3 or a multiple of 4 is given by

C ∪ D = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6), (1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2)}

⇒ n(C ∪ D) = 20.

Therefore, the probability that the sum of the numbers rolled is either a multiple of 3 or a multiple of 4 will be

$P(C\cup D)=\dfrac{n(C\cup D)}{n(S)}=\dfrac{20}{36}=\dfrac{5}{9}.$

Thus, the required probabilities are

(1) $\dfrac{11}{18}.$

(2) $\dfrac{5}{9}.$

Answer 2

The PLATO answer is for the first probability is 11/18 and the second probability is 5/9.

 

A pair of dice has 36 possible outcomes.

22 out of 36 have a sum of an even number or a multiple of 5 and when simplified is 11/18.

20 out of 36 have a multiple of 3 or 4 and when simplified is 5/9.