Pls help I'm sleep deprived rn and am almost done with this
1 answer:
You might be interested in
Answer:
1.B. No. You need to know the value of P(A and B). 2.C. Yes P(A and B) =0, so P(A or B) = P(A) + P(B).
Step-by-step explanation:
We can solve this question considering the following:
For two mutually exclusive events:
(1)
An extension of the former expression is:
(2)
In <em>mutually exclusive events,</em> P(A and B) = 0, that is, the events are <em>independent </em>one of the other, and we know the probability that <em>both events happen</em> <em>at the same time is zero</em> (P(A <em>and</em> B) = 0). There are some other cases in which if event A happens, event B too, so they are not mutually exclusive because P(A <em>and</em> B) is some number different from zero. Notice the difference between <em>OR</em> and <em>AND. The latter implies that both events happen at the same time.</em>
In other words, notice that the formula (2) provides an extension of formula (1) for those events that are not <em>mutually exclusive</em>, that is, there are some cases in which the events share the same probabilities in a way that these probabilities <em>must be subtracted</em> from the total, so those probabilities in common do not "inflate" the actual probability.
For instance, imagine a person going to a gas station and ask for checking both a tire and lube oil of his/her car. The probability for checking a tire is P(A)=0.16, for checking lube oil is P(B)=0.30, and for both P(A and B) = 0.07.
The number 0.07 represents the probability that <em>both events occur at the same time</em>, so the probability that this person ask for checking a tire or the lube oil of his/her car is:
P(A or B) = 0.16 + 0.30 - 0.07 = 0.39.
That is why we cannot simply add some given probabilities <em>without acknowledging if the events are or not mutually exclusive</em>, whereas we can certainly add the probabilities in question when we know that both probabilities are <em>mutually exclusive</em> since P(A and B) = 0.
In conclusion, knowing the events are mutually exclusive <em>does</em> provide <em>extra information</em> and we can proceed to simply add the probabilities of either event; thus, the answers are those in which <em>we need to previously know the value of P(A and B)</em>.