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The probability that event A occurs is 0.4, and the probabil
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The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55
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Originally posted by shrive555 on 26 Dec 2010, 17:47.
Last edited by abhimahna on 28 Mar 2018, 03:13, edited 1 time in total.
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Re: The probability that event A occurs is 0.4, and the probabil
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26 Dec 2010, 20:42
P(A or B) = P (A) + P(B)  p(a n b) 0.6= 0.4 + P(B)  0.25 P(B) = 0.45




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26 Dec 2010, 22:30
P(AandB) = pA + pB  p(AintersectionB) 0.6= 0.4 + p(B)  0.25 = 0.45



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Re: The probability that event A occurs is 0.4, and the probabil
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27 Dec 2010, 02:29
you can state both equations only if they are independent from each other ...



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Re: The probability that event A occurs is 0.4, and the probabil
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10 Aug 2013, 04:33
hirendhanak wrote: P(A or B) = P (A) + P(B)  p(a n b) 0.6= 0.4 + P(B)  0.25 P(B) = 0.45 Hi. Can u tell me wat is P(AandB)???? Please clear my doubt. regards, Rrsnathan.



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Re: The probability that event A occurs is 0.4, and the probabil
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10 Aug 2013, 04:59
rrsnathan wrote: hirendhanak wrote: P(A or B) = P (A) + P(B)  p(a n b) 0.6= 0.4 + P(B)  0.25 P(B) = 0.45 Hi. Can u tell me wat is P(AandB)???? Please clear my doubt. regards, Rrsnathan. P(A and B)= probability both events (A,B) occur= P(A)*P(B). Hope this clear your doubt



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Re: The probability that event A occurs is 0.4, and the probabil
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10 Aug 2013, 13:10
shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? 0.05 0.15 0.45 0.50 0.55 P(A) = .40 P(A and B) = P(A) * P(B) = .25 1 P(A or B) = P (A) + P(B) = .60 2 Both 1 & 2 give different results. 1 => P(B) = \(.25/P(A)\) = 0.62 2=> P(B) = .60  .40 = .20 whats going wrong ...i don;t know The only correction you need is to minus... P(A or B) = P (A) + P(B)  P(A) * P(B) = or, 0.60 = 0.4 +p(B)  0.25 so, p(B) = 0.45 (C) we all make mistake....don't worry about it...



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Re: The probability that event A occurs is 0.4, and the probabil
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12 Oct 2017, 19:30
shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 P(A) = .40 P(A and B) = P(A) * P(B) = .25 1 P(A or B) = P (A) + P(B) = .60 2 Both 1 & 2 give different results. 1 => P(B) = \(.25/P(A)\) = 0.62 2=> P(B) = .60  .40 = .20 whats going wrong ...i don;t know If A and B are overlapping events then P (A or B) = P(A) + P(B)  P(A and B) C



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Re: The probability that event A occurs is 0.4, and the probabil
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12 Oct 2017, 19:36
shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 P(A) = .40 P(A and B) = P(A) * P(B) = .25 1 P(A or B) = P (A) + P(B) = .60 2 Both 1 & 2 give different results. 1 => P(B) = \(.25/P(A)\) = 0.62 2=> P(B) = .60  .40 = .20 whats going wrong ...i don;t know Another way of looking at this is that both events are overlapping events A and B can occur at the same time  this is basically the same as Total= A + B both > is the same thing as saying P(A or B)= P(A) + P(B) P(A and B) This is one application of the overlapping set formula to probability



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The probability that event A occurs is 0.4, and the probabil
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Updated on: 08 Jun 2018, 12:00
Nunuboy1994 wrote: shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur? A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 P(A) = .40 P(A and B) = P(A) * P(B) = .25 1 P(A or B) = P (A) + P(B) = .60 2 Both 1 & 2 give different results. 1 => P(B) = \(.25/P(A)\) = 0.62 2=> P(B) = .60  .40 = .20 whats going wrong ...i don;t know Another way of looking at this is that both events are overlapping events A and B can occur at the same time  this is basically the same as Total= A + B both > is the same thing as saying P(A or B)= P(A) + P(B) P(A and B) This is one application of the overlapping set formula to probability I have a similar question as shrive555. Not really convinced with Nunuboy1994's explanation. shrive555 When events are mutually exhaustive, then P(A or B) = P(A) + P(B). So we can't use that formula (i.e. your point 2) since the Q tells us that both events can happen together. Hey abhimahna Here's my question: How do we know that the events are not independent? For independent events, we know that P(A and B) = P(A) * P(B), which would give a different answer.
Originally posted by dabaobao on 04 Jun 2018, 17:15.
Last edited by dabaobao on 08 Jun 2018, 12:00, edited 1 time in total.



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The probability that event A occurs is 0.4, and the probabil
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04 Jun 2018, 22:48
Alternate approach P(Total) = 1  P(Event A) = 0.4  P(Both) = 0.25 (from question stem) P(Neither) = 1  P(Either event A or event B) = 1  0.6 = 0.4 P(Total) = P(Event A) + P(Event B)  P(Both) + P(Neither)Substituting values, \(1 = 0.4 + P(Event B)  0.25 + 0.4\) > \(1 = 0.8  0.25 + P(Event B)\) > \(P(Event B) = 1  0.55\) = 0.45(Option C)
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The probability that event A occurs is 0.4, and the probabil
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Updated on: 15 Jun 2018, 10:30
pushpitkc wrote: Alternate approach
P(Total) = 1  P(Event A) = 0.4  P(Both) = 0.25 (from question stem)
P(Neither) = 1  P(Either event A or event B) = 1  0.6 = 0.4
P(Total) = P(Event A) + P(Event B)  P(Both) + P(Neither)
Substituting values, \(1 = 0.4 + P(Event B)  0.25 + 0.4\) > \(1 = 0.8  0.25 + P(Event B)\) > \(P(Event B) = 1  0.55\) = 0.45(Option C) Thanks pushpitkc for that alternate approach. We could do that if we know that we have overlapping sets. Hey abhimahna, any idea what's the relationship between overlapping and independent sets? Why can't we use the independent events formula here, P(A and B) = P(A) * P(B), which would give a different answer.
Originally posted by dabaobao on 05 Jun 2018, 04:00.
Last edited by dabaobao on 15 Jun 2018, 10:30, edited 1 time in total.



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Re: The probability that event A occurs is 0.4, and the probabil
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06 Jun 2018, 15:42
shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur?
A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 We can use the formula: P(A or B) = P(A) + P(B)  P(A and B) So we have: 0.6 = 0.4 + P(B)  0.25 0.6 = 0.15 + P(B) 0.45 = P(B) Answer: C
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Re: The probability that event A occurs is 0.4, and the probabil
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15 Jun 2018, 10:44
dabaobao wrote: Hey abhimahna, any idea what's the relationship between overlapping and independent sets? Why can't we use the independent events formula here, P(A and B) = P(A) * P(B), which would give a different answer. Hey dabaobao , When two events are said to be independent of each other, what this means is that the probability that one event occurs in no way affects the probability of the other event occurring. Example  Say you rolled a die and flipped a coin. Here, no such case is happening. We have two events A and B as well as we have an overlap between them as well Both A and B. Hence, we will use the formula mentioned above. Also, please note that I would not consider A and B independent events unless explicitly stated in the question. Does that make sense?
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Re: The probability that event A occurs is 0.4, and the probabil
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15 Jun 2018, 11:18
0.6 = 0.4 + PB  0.25 PB = 0.45 Answer D



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Re: The probability that event A occurs is 0.4, and the probabil
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21 Jul 2018, 07:28
shrive555 wrote: The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur?
A. 0.05 B. 0.15 C. 0.45 D. 0.50 E. 0.55 Using Manhattan book formula
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Re: The probability that event A occurs is 0.4, and the probabil
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22 Dec 2019, 22:13
Please forgive my ignorance, but why P(B) not equal to 0.625?? as P(A) = .40 P(A and B) = P(A) * P(B) = .25 P(B) = .25/P(A) = 0.625
I am deeply confused here! Please help me!



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Re: The probability that event A occurs is 0.4, and the probabil
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22 Dec 2019, 22:18
pushpitkc wrote: Alternate approach
P(Total) = 1  P(Event A) = 0.4  P(Both) = 0.25 (from question stem)
P(Neither) = 1  P(Either event A or event B) = 1  0.6 = 0.4
P(Total) = P(Event A) + P(Event B)  P(Both) + P(Neither)
Substituting values, \(1 = 0.4 + P(Event B)  0.25 + 0.4\) > \(1 = 0.8  0.25 + P(Event B)\) > \(P(Event B) = 1  0.55\) = 0.45(Option C) Please forgive my ignorance, but why P(B) not equal to 0.625?? as P(A) = .40 P(A and B) = P(A) * P(B) = .25 P(B) = .25/P(A) = 0.625 Secondly, can you please explain what do you mean by "1  P(Event A) = 0.4  P(Both)" =0.25? I am not familiar with why "total 1" sign "" P(A)=0.4 sign "" P(Both)? I am deeply confused here! Please help me!



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Re: The probability that event A occurs is 0.4, and the probabil
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22 Dec 2019, 23:01
gaoyuskr wrote: pushpitkc wrote: Alternate approach
P(Total) = 1  P(Event A) = 0.4  P(Both) = 0.25 (from question stem)
P(Neither) = 1  P(Either event A or event B) = 1  0.6 = 0.4
P(Total) = P(Event A) + P(Event B)  P(Both) + P(Neither)
Substituting values, \(1 = 0.4 + P(Event B)  0.25 + 0.4\) > \(1 = 0.8  0.25 + P(Event B)\) > \(P(Event B) = 1  0.55\) = 0.45(Option C) Please forgive my ignorance, but why P(B) not equal to 0.625?? as P(A) = .40 P(A and B) = P(A) * P(B) = .25 P(B) = .25/P(A) = 0.625 Secondly, can you please explain what do you mean by "1  P(Event A) = 0.4  P(Both)" =0.25? I am not familiar with why "total 1" sign "" P(A)=0.4 sign "" P(Both)? I am deeply confused here! Please help me! Hi gaoyuskrProbability formula: Event (A OR B) or P (A U B) where P (A ∩ B) = P(A and B) P (A U B) = P (A) + P (B) – P (A ∩ B) Using values: .6 = .4 + P(B)  .25 P(B) = .45 Option C! Hope it is clear.
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