Events A, B, and C are defines on sample space S. Their corresponding sets of sample points do not intersect and their union is S. Further, event B is twice as likely to occur as event A, and event C is twice as likely to occur as event B. Determine the probability of each of these three events.
Given:
S=A∪B∪C=sample space
∅=A∩B
∅=B∩C
∅=C∩A
P(A)=2P(B)
P(B)=2P(C)
Therefore
P(S)=1
P(A)+P(B)+P(C)=1
2P(B)+2P(C)+P(C)=1
4P(C)+2P(C)+P(C)=1
7P(C)=1
P(C)=1/7
Can you find the rest?
To determine the probability of each event, we need to make use of the given information and apply some probability principles.
Let's start by assigning probabilities to the events A, B, and C.
Let P(A) be the probability of event A occurring.
Since event B is twice as likely to occur as event A, we can write P(B) = 2P(A).
Similarly, event C is twice as likely to occur as event B, so we have P(C) = 2P(B).
The total probability of the sample space S is the sum of the probabilities of each event, which is equal to 1:
P(S) = P(A) + P(B) + P(C) = 1
We also know that the corresponding sets of sample points for events A, B, and C do not intersect and their union is S. This means that the events are mutually exclusive.
So, the probability of the sample space S is the sum of the probabilities of each event:
P(S) = P(A) + P(B) + P(C) = 1
Now, let's substitute the values of P(B) and P(C) in terms of P(A):
P(S) = P(A) + 2P(A) + 2(2P(A)) = 1
Simplifying the equation:
P(S) = P(A) + 2P(A) + 4P(A) = 1
7P(A) = 1
P(A) = 1/7
Using this value of P(A), we can calculate the probabilities of events B and C:
P(B) = 2P(A) = 2(1/7) = 2/7
P(C) = 2P(B) = 2(2/7) = 4/7
Therefore, the probability of event A is 1/7, the probability of event B is 2/7, and the probability of event C is 4/7.