Question

A certain experiment has only A, B, and C as the possible outcomes. If A and B are mutually exclusive events and P(A or B) = 0.7. If B and Care independent events and P (B and C) = 0.2, P(A) = ?​

Answer 1

Step-by-step explanation:

A certain experiment has three possible outcomes. The outcomes are mutually exclusive and have probabilities p, , and , respectively. What is the value of p?

A)

B)

C)

D)

E)

Mutually exclusive - In probability , two mutually exclusive or disjoint events MEANS both cannot occur together. example : a single coin toss, which can result in either heads or tails, but not both.

So this means there is no interaction of the 3 events and also there are only these three outcomes possible..

So addition of their probabilities should equal 1..

So

Answer 2

Answer:

1/30

Step-by-step explanation:

Since the experiment has only 3 events, therefore the sum of there probabilities should be 1.

P(A) + P(B) + P(C) = 1 ---------- Eq 1

Events A and B are mutually exclusive, therefore

P(A) + P(B) = 0.7 --------------- Eq 2

Events B and C are independent events, therefor

P(B) * P(C) = 0.2 ---------------- Eq 3

On substituting Eq 2 in Eq 1, we get

0.7 + P(C) = 1

P(C) = 0.3

Now replace the value of P(C) in Eq 3

P(B) * 0.3 = 0.2

P(B) = 2/3

Now replacing the value of P(B) in equation 2, we get

P(A) = 7/10 - 2/3

P(A) = 1/30 and that is the answer.