There are two types of guys "Duds" and "Dudes".
Let probability of a date with "Dud" be P(dud) and date with dude is P(dude).

Do P(dud)+P(dude)=1. Since a guy can either be a dud or a dude.

Now there were 2 duds in a row. So probability of that is P(dud)*P(dud)= 4/25.

Hence P(dud)= 2/5. And from the first equation we have P(dude)=1-P(dud)= 3/5.

Now if the question is asking what is the chance of getting 3 duds in a row then the probability should be P(dud)^3 or 8/125.
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Although not stated, the assumption here is that, on ANY GIVEN DATE, the probability of getting a dud is always the same.
Let y = P(getting a dud on a given date)

We're told that P(2 duds in a row) = 4/25
In other words: P(dud on 1st date AND dud on 2nd date) = 4/25
Rewrite as: P(dud on 1st date) X P(dud on 2nd date) = 4/25
So, we get: y X y = 4/25
In other words, y² = 4/25
Solve to get: y = 2/5

So, P(getting a dud on a given date) = 2/5
So, P(NEXT date is a dude) = 1 - 2/5 = 3/5

Answer:

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Hi dear Sandy,
Your solution is correct but in order to clarify my own concepts regarding probabilities I have some queries regarding this question as probability questions mostly seem confusing at first glance.
The following things are deductible from the question statement:
i). Both the events of a date being either a dud or a dude are independent
ii). There is no mention of both the events being equally likely though it may be implied but no surety
ii). Both the events are not mutually exclusive

I have following queries
i). Why have you assumed that both the events of next date being a dud or a dude are equally likely i.e. they both have same probability like tossing of a fair coin?
ii). Probability of a dud could be P(dud)= 1/5 instead of P(dud)= 2/5 as both the events are independent so for both events to occur in a row P(dud)*P(dude)= (2/5)*(2/5)= (1/5)*(4/5)= 4/5
=> if P(dud)= 1/5 and P(dud)+P(dude) = 1 => P(dude) = 1-(1/5) = 4/5

regards
m.yasir

Hi Yasir9909,

Is the word above (in blue) correct?

Cheers,
Brent
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@ GreenlightTestPrep

Yes it is correct; dud means 'something that has no value or that does not work'.
Your explanation of question sounds quite reasonable.

regards
m.yasir

I ask because you have indicated that the correct answer is D (3/5). This suggested to me that perhaps you meant to ask for P(dude), which would be 3/5
However, as it stand, the correct answer is C (2/5)

Cheers,
Brent
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I just checked, and this is a Princeton Review question - https://books.google.ca/books?id=9VgwAA ... er&f=false

The question asks for P(dude) not P(dud)

Cheers,
Brent
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Hi Brent,

You are right;sorry it was typo mistake on my part,it was actually probability of dude that we are required to calculate.
Thanks for correction.I could not understand you query at first.Depth of your knowledge and your skills in mathematics are quite impressive

regards
m.yasir

I've edited my initial response to match the edited question.

Cheers,
Brent
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Please guys.

When you post a new one question follow the rules for posting.....

At least 3 TAGS.

Thank you

Hi,


I do not see any option,how to use tag...

Concept has become clear to me now.Here is how:
Probability of two duds in a row = 4/25

Since Probability of 1st Dud is independent of probability probability of second dud and same goes with probability of second dud in relation with having first dud.
Apart from this Probability of first dud = Probability of 2nd dud

=> P(1st dud) = P(2nd Dud) = P(Dud) = x ;assume both of them to be equal x

P(1st dud) and P(2nd Dud) are independent so P(1st dud)*P(2nd Dud) = x^2 = 4/25

=> P(Dud) = x = 2/5
also P(Dud)+P(Dude) = 1

=> P(Dude) = 1 -P(Dud) = 1 - 2/5 = 3/5

so Probability of having a dude on next date is : P(Dude) = 3/5

I picked answer E. Can someone tell me where I went wrong?
P(dud)*p(dud) = 4/25
Since certain events have already happened, i.e Portia's first two dates occurred in the past, we can assign them a probability of 1 (it is 100% likely that they happened since we are told these events definitely happened.)
P(dud)(date 1) =1
P(dud) (date 2) = 1
We can ignore the first date since we are only looking at the probability of Portia's next date.
P(dud)(date2)*P(dud)(date 3) = 4/25.
1*P(dud)(date 3)=4/25
P(dud)(date 3)=4/25
p(dude)(date 3) = 1-4/25
p(dude)(date 3)= 21/25

I am ok with this explanation but since in the question is clearly specified that the second dud is different from the first one how can it be equally likely that the guy she dates is a dud? Assuming there is a certain number X of dud guys and N is the total number of guys, shouldn't this probability be \(\frac{X}{N}*\frac{X-1}{N-1}\). It is as there is no reimmision. Am I wrong?

Why not E

The date on Tuesday is irrelevant, on Wednesday night he was dud on Thursday he could be a dud or dude, the chance that he would be a dud is 4/25 because if he is a dud she have scored dating two dud conservatively, so the chance that he would be a dude is 1-4/25 = 21/25... I think E is the answer.

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