Answer to Question #212836 in Statistics and Probability for kojo

Question

 

 

 

     6.    (a)      Gideon rolls two (2) six-sided dice 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). Probability that the sum is even.

The sample space of rolling two, six-sided dice and summing the outcomes is shown in the table below.

Let O represent odd and D represent divisible by 5.

$P(O\cup D)=P(O)+P(D)-P(O\cap D)$

From the table, 18 out of 36 are odd, thus ) $P(O)=0.5$

Similarly, there are 7 out of 36 numbers that are divisible by 5. Thus, $P(D)=\frac{7}{36}$

3 numbers (5s) are both odd and divisible by 5. Thus, $P(O\cap D)=\frac{1}{12}$

Therefore,

$P(O\cup D)=\frac{1}{2}+\frac{7}{36}-\frac{1}{12}=\frac{11}{18}$

Answer $\frac{11}{18}$

(b). $P(6|DI)$

Let DI represent different numbers and 6 represent at least one landing on 6.

$P(6|DI)=\frac{P(6\cap DI)}{P(DI)}$

Probability of at least one landing on 6 and having different numbers is the probability of one landing on 6 and not both, represented by numbers in the 6th row and 6th column except the diagonal value (12). 5 for one dice and five fore the second.

$P(6\cap DI)=\frac{5}{36}+\frac{5}{36}=\frac{5}{18}$

Having different outcomes is represented by off-diagonal numbers in the table. Thus,

$P(DI)=\frac{30}{36}=\frac{5}{6}$

$P(6|DI)=\frac{\frac{5}{18}}{\frac{5}{6}}=\frac{1}{3}$

Let me answer it using a different version.

Let F be the dice landing on different numbers.

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

$p(F)=\frac{30}{36}=\frac{5}{6}$

Let E be at least one dice landing on 6.

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

$E\cap F = {(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)}$

Thus, $P(E\cap F)=\frac{10}{36}=\frac{5}{18}$

$P(E|F)=\frac{P(E\cap F}{P(F)}$

$=\frac{5}{18}×\frac{6}{5}=\frac{1}{3}$