Respuesta :
Answer:
The probabilities are [tex]\dfrac{2}{15}[/tex] and [tex]\dfrac{4}{5}[/tex].
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.
[tex]\text{Probability}=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
Thus,
[tex]P(A)=\frac{n(A)}{n(S)}=\frac{5}{36}[/tex]
[tex]P(B)=\frac{n(B)}{n(S)}=\frac{30}{36}=\frac{5}{6}[/tex]
[tex]P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{4}{36}=\frac{1}{9}[/tex]
Hence, the probability that the sum is 6 if the number are different,
[tex]P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}[/tex]
[tex]=\frac{\frac{1}{9}}{\frac{5}{6}}[/tex]
[tex]=\frac{6}{45}[/tex]
[tex]=\frac{2}{15}[/tex]
Similarly,
The probability that the numbers are different if the sum is 6,
[tex]P(\frac{B}{A})=\frac{P(A\cap B)}{P(A)}[/tex]
[tex]=\frac{\frac{1}{9}}{\frac{5}{36}}[/tex]
[tex]=\frac{36}{45}[/tex]
[tex]=\frac{4}{5}[/tex]
