Respuesta :
Answer:
The statement which is correct is:
B. A and B cannot be both independent and mutually exclusive.
Step-by-step explanation:
Let we toss a coin single time.
and event A denote the tossing up a head.
and event B denote tossing a tail.
- Now, we know that two events A and B are said to be mutually exclusive if:
P(A∩B)=0
- and if two events A and B are independent then:
P(A∩B)=P(A)×P(B)
Here we have:
P(A)≠0 and P(B)≠0
So, if the events are independent then we have:
P(A∩B)≠0
Hence, the events can't be mutually exclusive.
Similarly if the two events are mutually exclusive then they can't be independent.
i.e. both mutually exclusive property and independent property can't exist at the same time if the two events have non-zero probability.
Hence, the answer is: Option: B
Using probability concepts, it is found that the correct option is:
B. A and B cannot be both independent and mutually exclusive
If two events, A and B, are independent, we have that:
[tex]P(A)P(B) = P(A \cap B)[/tex]
If they are mutually exclusive, we have that:
[tex]P(A \cap B) = P(A)P(B|A) = 0[/tex]
In which P(B|A) is the probability of B happening when A has happened.
Considering both P(A) and P(B) are not zero, if they are independent:
[tex]P(A \cap B) = P(A)P(B) \neq 0[/tex]
Since [tex]P(A \cap B) \neq 0[/tex], they cannot be both independent and mutually exclusive, thus, the correct option is:
B. A and B cannot be both independent and mutually exclusive
A similar problem is given at https://brainly.com/question/14478923
