Answer:
a. [tex]\dfrac{1}{4}[/tex]
b. [tex]\dfrac{1}{12}[/tex]
c. Not mutually exclusive.
d. Not independent events
Step-by-step explanation:
a. When two six-sided fair dice are rolled one time,
Total possible outcomes, n(S) = 6 × 6 = 36,
If A be the event that the first die is odd and the second is a 4, 5, or 6
Then, A = {(1, 4), (3, 4), (5, 4), (1, 5), (3, 5), (5, 5), (1, 6), (3, 6), (5, 6)}
i.e. n(A) = 9,
[tex]\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]
Thus, the probability of event A,
[tex]P(A) = \frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}[/tex]
b. if B be the event that the sum of the two dice is 10,
Then B = {(4, 6), (5, 5), (6, 4)}
i.e. n(B) = 3,
Thus, the probability of event B,
[tex]P(B) = \frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}[/tex]
c. Two event are called mutually exclusive events,
If they are disjoint,
i.e. A and B are mutually exclusive,
If A ∩ B = ∅
∵ {(1, 4), (3, 4), (5, 4), (1, 5), (3, 5), (5, 5), (1, 6), (3, 6), (5, 6)} ∩ {(4, 6), (5, 5), (6, 4)}
= {(5, 5)} ≠ ∅
So, they are not mutually exclusive events.
d. Two events are called independent if the occurrence of one event does not affect the occurrence of other event.
Also, A and B are independent events,
If P(A ∩ B) = P(A) × P(B)
n(A∩ B) = 1,
[tex]\implies P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{1}{36}[/tex]
∵ [tex]\frac{1}{4}\times \frac{1}{12}=\frac{1}{48}\neq \frac{1}{36}[/tex]
Hence, they are not independent events.