Answer to Question #218231 in Statistics and Probability for Bless

Question

a) Gideon rolls two(2) six-sided die once. What is the probability that the sum of the outcomes of both dice are odd or divisible by 5?

b) Two (2) dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on two different numbers?

Answer 1

a)

Sum table

There are 36 possible outcomes.

P(sum is odd) $= \frac{18}{36}$

Possible combinations when sum is divisible by 5: (1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)

Total number = 7

P(sum is divisible by 5) $= \frac{7}{36}$

Possible combinations when sum both odd and divisible by 5: (1,4), (2,3), (3,2), (4,1)

P(both odd and divisible by 5) $= \frac{4}{36}$

P(the sum of the outcomes of both dice are odd or divisible by 5) = P(sum is odd) + P(sum is divisible by 5) - P(both odd and divisible by 5)

$= \frac{18}{36} + \frac{7}{36} -\frac{4}{36} \\ = \frac{21}{36} \\ = \frac{7}{12}$

b)

A = Dice land on two different numbers

A = {(1.2), (1,3), (1,4), (1,5), (1,6),

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

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

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

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

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

n(A)=30

B = At least one lands on 6

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

n(B)=11

$A \cap B =$ {(1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5)}

$n(A \cap B) =10 \\ P(\frac{B}{A)}= \frac{P(A \cap B)}{P(A)} \\ = \frac{\frac{10}{36}}{\frac{30}{36}} \\ = \frac{10}{30} \\ = \frac{1}{3}$