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Eventsindependent: Suppose een Occursprobability 080 and avcnt occurs with probability 0.26Cometetha nrobabibty thab Dccuts but notGecur Compute the probability tha...

Question

Eventsindependent: Suppose een Occursprobability 080 and avcnt occurs with probability 0.26Cometetha nrobabibty thab Dccuts but notGecur Compute the probability that cither Ocoim wihout accumingboth Oerurncccssan consult 0 Est_ot formulas )

Events independent: Suppose een Occurs probability 080 and avcnt occurs with probability 0.26 Cometetha nrobabibty thab Dccuts but notGecur Compute the probability that cither Ocoim wihout accuming both Oerur ncccssan consult 0 Est_ot formulas )



Answers

If $P(A)=0.3$ and $P(B)=0.4$ and $A$ and $B$ are independent events, what is the probability of each of the following? a. $\quad P(A \text { and } B)$ b. $\quad P(\mathbf{B} | \mathbf{A})$ c. $\quad P(\mathrm{A} | \mathrm{B})$

Okay, so we consider for independent events one eight Soon, 34 And we went p I You could see the probability of one of the eyes and last expressed a probability that at least one of the four events occur in terms of P eyes. So at least one of the same as one minus probable ability of none of them occurring that's equal to sort of probable date of them occurring. Is this equal to he? I So that's one minus the 1st 1 on a great second. We're not a great they're not occurring and forth will not agree. Get it past the probability. Uh, at least two of the events occurring they're not equipped to one minus the probability of mentoring or the probability of Wonder cream. What is that equal to where we found the probability of, um, none. Which is this thing over here? That's one minus. You're not one of this probably would have done is just this portion Here. It's one when it's p one. Well, I'm gonna speak to one minus p three and one win is Pete for plus the probability of just wanna cry. So that's either of you wanting to court p two concurrent he drink maker or P for kicker

In this problem, we are going to use the multiplication rule of probability. And in the first part of the question is given the P L E Is equal to 0.2 And the probability of the event, BP of B is 0.4. Now considering A and B to be independent events, what will be the value of P A and B? So if A and B are independent events, then the probability of E and B. Using the multiplication rule is simply P L A times B R B. So P of a 0.2, M P S V 0.4. So the probability of A and me will be point to transpoint four, which is equal to 0.8 Now, in the second part of the question and fired me, it is given that P L. E given me is equal to 0.1. And what we need to do is compute D of E. And now P of given me is equal to B, O E, and B divided by the F B. So this will be equal to 0.1. So this implies that P M E, and B Divided by the value of TV is given to be 0.4. So we will substitute that over here, and this will be equal to 0.1. So from here we can just do a simple cross multiplication and obtain the PLE. And B is equal to 0.1 and 0.4 and 0.1 times 0.4. Z equal to 0.4 So the probability of A and me that is P of A and B is equal to 0.4 Yeah.

All right. So for number 26 were given the probability of event one and two or 1/4 and one half, and we want to see the probability of both of them occurring if they're independent. So that's when you multiply 1/4 times one half to get your final answer of.

This problem requires us to use a multiplication rule when two events and be our independent, the probability that both occur. So the probability that a and B occur is the probability of a multiplied by the probability of B and this problem. We're told that the probability of A and B occurring is 0.13 and the probability of a occurring is for tense. We're looking for the probability of B solving this algebraic lee. If I divide both sides by 0.4, I'll find that the probabilities that be occurs is 0.3 to 5.


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