Respuesta :
Answer:
Option 1 - No, the probability of F occurring is higher if it is known that E has occurred.
Step-by-step explanation:
Given : Roll a die and consider the following two events={2, 3, 6}, event ={1, 5, 6}.
To find : Are the events E and F independent?
Solution :
E={2, 3, 6}, F={1, 5, 6}
They are not independent event as 6 is common in both the events.
Now, verifying by applying independent property,
Probability of event E is [tex]P(E)=\frac{3}{6}=\frac{1}{2}[/tex]
Probability of event F is [tex]P(F)=\frac{3}{6}=\frac{1}{2}[/tex]
Probability of E and F is [tex]P(E\cap F)=\frac{1}{6}[/tex]
Probability of F occuring when it is known that e has occured.
i.e. [tex]P(F/E)=\frac{P(E\cap F)}{P(E)}[/tex]
[tex]P(F/E)=\frac{\frac{1}{6}}{\frac{1}{2}}[/tex]
[tex]P(F/E)=\frac{1}{3}[/tex]
[tex]P(F)>P(F/E)[/tex]
i.e. The probability of F occurring is higher if it is known that E has occurred.
Therefore, The correct option is 1.
Events E and F are not independent, because the probability of F occurring is higher if it is known that E has occurred.
Answer: The correct option is
(B) No, because there is at least one roll which leads to both E and F occurring.
Step-by-step explanation: We are given to roll a die and consider the following two events :
A = {2, 3, 6} and B = {1, 5, 6}.
We are to check whether the events A and B are independent.
We have
A ∩ B = {2, 3, 6} ∩ {1, 5, 6} = {6}.
So, n(A) = 3, n(B) = 3 and n(A ∩ B) = 1.
Since there are 6 elements in the sample space, we get
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6.
So, the probabilities of the events A, B and A ∩ B are calculated as follows :
[tex]P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{6}=\dfrac{1}{2},\\\\\\P(B)=\dfrac{n(B)}{n(S)}=\dfrac{3}{6}=\dfrac{1}{2},\\\\\\P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{1}{6}.[/tex]
Therefore,
[tex]P(A\cap B)=\dfrac{1}{6}\neq P(A)\times P(B)=\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4}[/tex]
Thus, the events A and B are NOT independent because at least one roll which leads to both E and F occurring and that is 6, common to both A and B.
Option (B) is CORRECT.
