Question

Benicio rolls a pair of number cubes. What is the probability that the sum is 6 if the number are different? What is the probability that the numbers are different if the sum is 6?

Answer 1

Answer:

The probabilities are $\dfrac{2}{15}$ and $\dfrac{4}{5}$.

Step-by-step explanation:

When a dice is rolled, then total possible outcomes = 6 (i.e. 1,2,3,4,5 or 6).

So, when a pair of dice is rolled,

Total possible outcomes, n(S) = 6 × 6 = 36

Consider A represents the event of getting a sum of 6 and B represents the event of different numbers appear.

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

B={(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)}

A∩B= {(1,5),(2,4),(4,2),(5,1)}

So, n(A) = 5, n(B) = 30, n(A∩B) = 4.

$\text{Probability}=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

Thus,

$P(A)=\frac{n(A)}{n(S)}=\frac{5}{36}$

$P(B)=\frac{n(B)}{n(S)}=\frac{30}{36}=\frac{5}{6}$

$P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{4}{36}=\frac{1}{9}$

Hence, the probability that the sum is 6 if the number are different,

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}$

         $=\frac{\frac{1}{9}}{\frac{5}{6}}$

         $=\frac{6}{45}$

         $=\frac{2}{15}$

Similarly,

The probability that the numbers are different if the sum is 6,

$P(\frac{B}{A})=\frac{P(A\cap B)}{P(A)}$

         $=\frac{\frac{1}{9}}{\frac{5}{36}}$

         $=\frac{36}{45}$

         $=\frac{4}{5}$