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The probability that event M will not occur is 0.8 and the probability
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15 Jun 2016, 00:32
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The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur? A) 1/5 B) 2/5 C) 3/5 D) 4/5 E) 12/25
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The probability that event M will not occur is 0.8 and the probability
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11 Oct 2016, 15:59
Bunuel wrote: The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur?
A) 1/5 B) 2/5 C) 3/5 D) 4/5 E) 12/25 We are given that the probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6 and that events M and R cannot both occur. We need to determine the probability that either event M or event R will occur. The probability that event M will occur is 1  0.8 = 0.2 = 1/5 The probability that event R will occur is 1  0.6 = 0.4 = 2/5 Since events M and R cannot both occur , the probability that either event M or event R will occur is 1/5 + 2/5 =3/5. Answer: C
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Re: The probability that event M will not occur is 0.8 and the probability
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15 Jun 2016, 03:04
Bunuel wrote: The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur?
A) 1/5 B) 2/5 C) 3/5 D) 4/5 E) 12/25 P(R) = 1  0.8= 0.2 P(M) = 1 0.6 = 0.4 Given that both events are mutually exclusive. Prob that either event M or event R will occur = 0.2+0.4 = 0.6 = (6/10) = (3/5) C is the answer.




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Re: The probability that event M will not occur is 0.8 and the probability
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09 Oct 2016, 23:53
shonakshi wrote: Can someone please explain why is it not 0.8*0.4+0.2*0.6 11/25 I think , you are trying to calculate probability by multiplying P(M)*P(~R) + P(R)*P(~M) . This is wrong . Formula is P(M or R) = P(M) + P(R)  P (M and R) => 0.2 + 0.4  0 = > 0.6 or 3/5 (Answer is C) .



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Re: The probability that event M will not occur is 0.8 and the probability
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03 Jul 2016, 13:34
Just made a mistake by multiplying 0.2 and 0.4, got \(\frac{2}{25}\) Note to self: "multiply" when there is an "AND", and "add" when there is an "OR" we should be adding 0.2+0.4 = 0.6 or \(\frac{3}{5}\)
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The probability that event M will not occur is 0.8 and the probability
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10 Oct 2016, 06:58
shonakshi wrote: Can someone please explain why is it not 0.8*0.4+0.2*0.6 11/25 Go through this page, it will answer all your queries... https://people.richland.edu/james/lectu ... 5rul.htmlHope that helps..
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The probability that event M will not occur is 0.8 and the probability
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Updated on: 15 Jun 2016, 03:31
p(m) =0.2 p(r) =0.4 p(m intersection r) = 0 (If events M and R cannot both occur) p(m or r) = 0.2+0.4 =0.6
Corrected !!
Originally posted by CounterSniper on 15 Jun 2016, 02:52.
Last edited by CounterSniper on 15 Jun 2016, 03:31, edited 1 time in total.



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Re: The probability that event M will not occur is 0.8 and the probability
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13 Oct 2016, 06:44
Bunuel wrote: The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6.If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur?
A) 1/5 B) 2/5 C) 3/5 D) 4/5 E) 12/25 M' = 0.80 So M = 0.20 R' = 0.60 So R = 0.40 Quote: probability that either event M or event R will occur? M+R = M + R  MR Quote: events M and R cannot both occur So , MR = 0 M+R = 0.20 + 0.40  0 Or, M+R = 0.60 Hence answer will be 0.60 or (C) 3/5 SOHAM6185 wrote: We should take into consideration that both cannot occurs. please explain The reason is highlighted... Try to solve this question using VENN Diagram approach it will be crystal clear...
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Re: The probability that event M will not occur is 0.8 and the probability
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26 Jan 2019, 09:31
Bunuel wrote: The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur?
A) 1/5 B) 2/5 C) 3/5 D) 4/5 E) 12/25 I will use the 2*2 matrix here MO= M will occur MNO= M will not occur All the values are given, and we need to find MOMNO RO00.40.4 RNO0.2 0.40.6 0.20.81 either event M or event R will occur, this will be again => 1  (both will occur), Since in probability P(happening) + P(not happening) = 1 10.4 =0.6 C
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Re: The probability that event M will not occur is 0.8 and the probability
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17 Nov 2019, 18:37
legendinthewomb wrote: chetan2u, Bunuel, VeritasKarishma, Gladiator59, generisA lot of users have tried explaining why
0.8*0.4+0.2*0.6 11/25 is wrong, but I think the confusion still standsWe understand why the correct soln is correct but we don't understand why our soln is incorrect. Can you help us bridge the gap? Thanks for your time and help Hi It is given that both events A and B will NOT occur together, so when you take one occurring say A =0.4, other B NOT occurring is not 0.8, it is 1 or 100% because it is sure that B will not occur when A is occurring. So answer is 0.2+0.4=0.6 Say you have only 20 cards of a pack of 52 cards. Probability of picking not picking an A is 0.9 and probability of not picking King is 0.8. Probability of picking any one will be straight sum of individual probabilities that is 0.1+0.2 because they cannot occur together.
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Re: The probability that event M will not occur is 0.8 and the probability
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15 Jun 2016, 22:07
Got it wrong.
p(m) =0.2 p(r) =0.4
So, I calculated probability as sum of:
i) m occurs but r does not occur = 0.2*0.6 ii) r occurs but m does not occur = 0.4*0.8
So, probability = 0.2*0.6 + 0.4*0.8 = 0.44 = 11/25!



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Re: The probability that event M will not occur is 0.8 and the probability
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26 Sep 2016, 17:55
Can someone please explain why is it not 0.8*0.4+0.2*0.6 11/25



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Re: The probability that event M will not occur is 0.8 and the probability
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08 Nov 2017, 16:00
shonakshi wrote: Can someone please explain why is it not 0.8*0.4+0.2*0.6 11/25 The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur? Let's break this down M will not occur = 0.8; M will occur = 0.2 R will not occur = 0.6; R will occur = 0.4 P(M or R) = P(M) + P(R)  P (M and R) P(M or R) = 0.2 + 0.4  0 P(M or R) = 0.6 = 6/10 = 3/5 Note: Multiplying occurrence that will not occur + Multiplying occurrence that will occur is not equal to either event M or event R will occur



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Re: The probability that event M will not occur is 0.8 and the probability
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24 Aug 2018, 07:07
GMATSkilled wrote: shonakshi wrote: Can someone please explain why is it not 0.8*0.4+0.2*0.6 11/25 The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur? Let's break this down M will not occur = 0.8; M will occur = 0.2 R will not occur = 0.6; R will occur = 0.4 P(M or R) = P(M) + P(R)  P (M and R) P(M or R) = 0.2 + 0.4  0 P(M or R) = 0.6 = 6/10 = 3/5 Note: Multiplying occurrence that will not occur + Multiplying occurrence that will occur is not equal to either event M or event R will occur How can that formula be correct? What if the problem had said the probability that M will not occur is .5, and the probability that R will not occur is .5. Then according to your formula, you'd have: P(M or R) = P(M) + P(R)  P(M and R) P(M or R) = .5 + .5  0 P(M or R) = 1 So if two mutually exclusive events each have a 1/2 probability of occurring, then the probability of at least one of them occurring is 1 (100% of the time?) That doesn't make any sense to me, so I'm not exactly sure why that formula is what you're supposed to apply. In a reallife scenario, that's like saying the Cardinals have a 40% chance to win the World Series, and the Yankees have a 20% chance to win, so the chance that one of them wins is 60% (since both of them cannot win)... Which doesn't make any sense. Again, what if you said the cardinals have an 80% chance to win and the Yankees have a 30% chance to win. So the chance that one of them wins is 110%? The language in the question doesn't seem to suggest that either M or N must occur... In order to interpret the question the way the answer seems to, I would think the language would have to say something like "Out of 100 trials, M did not occur 80% of the time and N did not occur 60% of the time; if N and M never occurred on the same trial, what's the probability that a randomly selected trial will have M OR N occurring?" In that case, when you already have a list of events that happened, you could apply the formula P(M or R) = P(M) + P(R)  P(M and R)... Take it back to the 50% example, if you know that M occurred 50% of the time and N occurred 50% of the time, and M and N cannot happen, then you know that at least M or N had to happen every time. But when you're talking about the probability of a future event, as the problem seems to suggest, it doesn't make sense to just add the probabilities. Can someone tell me if/how my reasoning is incorrect?



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Re: The probability that event M will not occur is 0.8 and the probability
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10 May 2019, 10:02
shonakshi wrote: Can someone please explain why is it not 0.8*0.4+0.2*0.6 11/25 This is not right as you are confusing between Event and Occurrence of the event. Your formula would be true if you try to find out, probability of EXACTLY one of M or R will occur, when the other WILL NOT occur. This is NOT same as 'probability of M or R to occur'. I will illustrate with an example. Scenario 1: You are tossing a coin 2 times. What's the probability of getting EXACTLY one head? Your above formula will work in this case. There are 2 occurrences of toss here. Also, P(H) = 1/2 and P(T) = 1/2. Get EXACTLY one head in two tosses = P(H).P(~T) + P(~H).P(T) = 1/2. [without formula: HT (valid), TH (valid), HH (invalid), TT (invalid). Hence : 2/4 = 1/2] Scenario 2: What is the probability of getting a head OR tail when you toss a coin? Note: here it is talking about only one occurrence of toss. Here the formula should be P(H or T) = P(H) + P(T)  P(H and T). P(H and T) is obviously 0. So, P(H or T) = 1/2 + 1/2 = 1 (it's a certainty). [now without formula: it is a guarantee in a coin toss that you will either get head or tail. so, probability = 1 ] This question in OP is the second scenario described above, hence your formula won't give you correct result. Hope it helps !!



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Re: The probability that event M will not occur is 0.8 and the probability
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13 Oct 2016, 04:01
ScottTargetTestPrep wrote: Bunuel wrote: The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur?
A) 1/5 B) 2/5 C) 3/5 D) 4/5 E) 12/25 We are given that the probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6 and that events M and R cannot both occur. We need to determine the probability that either event M or event R will occur. The probability that event M will occur is 1  0.8 = 0.2 = 1/5 The probability that event R will occur is 1  0.6 = 0.4 = 2/5 Since events M and R cannot both occur , the probability that either event M or event R will occur is 1/5 + 2/5 =3/5. Answer: C We should take into consideration that both cannot occurs. please explain



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Re: The probability that event M will not occur is 0.8 and the probability
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30 Jun 2017, 02:45
sb0541 wrote: shonakshi wrote: Can someone please explain why is it not 0.8*0.4+0.2*0.6 11/25 I think , you are trying to calculate probability by multiplying P(M)*P(~R) + P(R)*P(~M) . This is wrong . Formula is P(M or R) = P(M) + P(R)  P (M and R) => 0.2 + 0.4  0 = > 0.6 or 3/5 (Answer is C) . yes, that is why p(m and n) need to be zero.



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Re: The probability that event M will not occur is 0.8 and the probability
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11 Mar 2018, 19:57
shonakshi wrote: Can someone please explain why is it not 0.8*0.4+0.2*0.6 11/25 Once you say that event M and event R can not occur together , it means that both are mutually exclusive. P(M).P(R)=0 Posted from my mobile devicePosted from my mobile device



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Re: The probability that event M will not occur is 0.8 and the probability
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14 Mar 2018, 07:18
be careful there are 2 cases case 1 1= p (a happen)+ p (b happen) p (both a and b happen) +p (neither a nor b happen) . this is ven digram
case 2 1= p (a dose not happen)+ p(b dose not happen) p (neither a nor b happen)+ p (both a and b happen). this is ven diagram p ( only a happen) is in p (b dose not happen).
so, there is two scenario here.



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Re: The probability that event M will not occur is 0.8 and the probability
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14 Sep 2018, 17:49
Can someone please address the above challenge to this question's answer? I also disagree with the application of the formula used in the solution. This question is posing a future probability event.
What if the question was modified to probability of M occurring=90% and R occurring=90%? (So not occurring for either event is 10%). M and R are still mutually exclusive events.
If we apply the recommended formula P(M or R)=P(M) + P(R) P(M and R), we get .9+.90=1.8. Wait, 1.8!? That doesn't even make sense.




Re: The probability that event M will not occur is 0.8 and the probability
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