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

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!

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 \(\frac{3}{5}\)

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) .

Go through this page, it will answer all your queries...

https://people.richland.edu/james/lectu ... 5-rul.html

Hope that helps..

M' = 0.80 So M = 0.20
R' = 0.60 So R = 0.40
M+R = M + R - MR
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


The reason is highlighted...

Try to solve this question using VENN Diagram approach it will be crystal clear...

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 real-life 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?

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

--------MO--------MNO
-RO----0----------0.4-----0.4
RNO---0.2--------0.4-----0.6
-------0.2---------0.8------1

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
1-0.4 =0.6

C

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 !!

chetan2u, Bunuel, VeritasKarishma, Gladiator59, generis

A 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 stands

We 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.

Solution attached

Probability of happening of an event + Probability of not happening of an event = 1

=> P(E) + P(E') = 1

=> P(M') = 0.8 and therefore P(M) = 1 - 0. 8 = 0.2

=> P(R') = 0.6 and therefore P(R) = 1 - 0. 6 = 0.4

P(MUR) = P(M) + P(R) - P(M ∩ R)

Given that events M and R cannot both occur: This is a mutually exclusive condition and hence P(M ∩ R) = 0

=> P(MUR) = P(M) + P(R)

=> P(MUR) = 0.2 + 0.4

=> P(MUR) = 0.6

=> P(MUR) = \(\frac{6 }{ 10}\) = \(\frac{3 }{ 5}\)

Answer C

When is it assumed that there can not be any other event except M or R?
Why is it assumed that other event can not occur either with M or R?

My approach:

Assume: A any event

I did as :
M+R+ A+ MA+RA+MR= 1
MR= 0
so , M+R+A+MA+RA = 1

Given"
R+A+RA= 0.8
M+A+MA= 0.6

Solving above 3 equations, we get
A= 1/5( any event can occur)

To find wither M or R occurs is
M+R+ MA+RA
it means 1-A
i.t. 4/5
Hence i am wrong

please suggest VeritasKarishma IanStewart MathRevolution ScottTargetTestPrep

I'm afraid I don't follow what you've done (I don't know what 'A' represents). We're not assuming in this question that no other event can occur. We just don't care at all about other events that might occur, because the question doesn't ask about anything besides M and R.

In this problem, the probability M occurs is 0.2, and the probability R occurs is 0.4. The question tells us they cannot both occur. That's the definition of 'mutually exclusive' events, and when events are mutually exclusive (can't both happen), we can add probabilities if we want to know if either event occurs. So 0.2 + 0.4 = 0.6 is the answer. It's just like this situation: in a bag, 20% of marbles are magenta-coloured (M), and 40% of marbles are red (R). If you pick one marble randomly, what is the probability it is magenta or red? It's just 60%, by adding. Here other events 'A' could happen -- maybe some marbles are large, some are small, and A is 'you pick a large marble' -- but we don't care about those events at all because the question doesn't ask about them.

Hi All,

This prompt tells us three Probability-based pieces of information:
1) The probability that Event M will NOT occur is 0.8
2) The probability that Even R will NOT occur is 0.6
3) Events M and R CANNOT BOTH occur.

We’re asked for the probability that Event M OR Event R will occur. While these types of probability questions are generally a bit harder in terms of difficulty, this particular prompt includes “causality” (which is a really rare concept in the Quant section) that actually makes the question easier to solve.

Normally, individual probabilities have NO impact on one another, but in this prompt, we are told that if Event M happens, then Event R CANNOT happen (meaning that there is an ‘absolute’ here and nothing to calculate). The same situation occurs if Event R happens (re: Event M automatically does NOT happen and there’s nothing to calculate).

Thus, there are just a couple of simple calculations to work through:

-The probability that Event M happens is 1 – 0.8 = 0.2
-The probability that Event R happens is 1 – 0.6 = 0.4

Thus, the probability that EITHER will happen is 0.2 + 0.4 = 0.6

Final Answer:
GMAT Assassins aren’t born, they’re made,
Rich