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On Saturday morning, Malachi will begin a camping vacation
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02 Sep 2010, 12:59
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On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday?
On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday? A. 0.008 B. 0.128 C. 0.488 D. 0.512 E. 0.640
We are looking for the probability of the following even NNR: no rain on first day, no rain on second day, rain on third day (Monday).
As the probability of rain on each day is 0.2 then the probability of not raining on each day is 1-0.2=0.8. So the probability of not raining on first and second days and raining on third day would be \(P(NNR)=0.8*0.8*0.2=0.128\).
Re: On Saturday morning, Malachi will begin a camping vacation
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12 Apr 2017, 12:10
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udaymathapati wrote:
On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday?
A. 0.008 B. 0.128 C. 0.488 D. 0.512 E. 0.640
Since we need to determine the probability that Malachi will return home at the end of the day on the following Monday, we must determine:
P(no rain Sat and no rain Sun and rain Mon) = P(no rain Sat) x P(no rain Sun) x P(rain Mon)
Since the probability of rain is 0.2, the probability of no rain is 1 - 0.2 = 0.8, and thus:
P(no rain Sat) x P(no rain Sun) x P(rain Mon) = 0.8 x 0.8 x 0.2 = 0.128
As the probability of rain on each day is 0.2 then the probability of not raining on each day is 1-0.2=0.8. So the probability of not raining on first and second days and raining on third day would be \(P(NNR)=0.8*0.8*0.2=0.128\).
Answer: B.
Hope it's clear.
Hey i understand this.. but i did not get the following line "Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains" what is this first day means here..
As the probability of rain on each day is 0.2 then the probability of not raining on each day is 1-0.2=0.8. So the probability of not raining on first and second days and raining on third day would be \(P(NNR)=0.8*0.8*0.2=0.128\).
Answer: B.
Hope it's clear.
Hey i understand this.. but i did not get the following line "Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains" what is this first day means here..
Anyway thanks for your time..
/Prabu
Vacation starts on Saturday and Malachi will return at the first day on which it rains. Question asks what is the probability that Malachi will return on next Monday, or what is the probability that it will rain on Monday (and not on Saturday or Sunday).
Hey i understand this.. but i did not get the following line "Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains" what is this first day means here..
Anyway thanks for your time..
/Prabu[/quote]
Vacation starts on Saturday and Malachi will return at the first day on which it rains. Question asks what is the probability that Malachi will return on next Monday, or what is the probability that it will rain on Monday (and not on Saturday or Sunday).
Bunuel, the question says ... returns on the following monday. Following monday means the next immediate monday or the monday after this one? Well, I took it to be the monday which comes in the week after. Guess "following" means the next. Anyways... got to be careful.
Re: On Saturday morning, Malachi will begin a camping vacation
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06 Nov 2016, 07:54
Bunuel wrote:
prabu wrote:
can you explain it?
/Prabu
As the probability of rain on each day is 0.2 then the probability of not raining on each day is 1-0.2=0.8. So the probability of not raining on first and second days and raining on third day would be \(P(NNR)=0.8*0.8*0.2=0.128\).
Answer: B.
Hope it's clear.
Aha, got it now, thanks Bunuel. Simpler than the convoluted steps I took. In my haste, I overlooked that the question was asking what the probability would be to return home on Monday only. I interpreted it as determine the probability of him returning on any of those days. If the question asked that, would this be correct? (A twist on the original question)(And I guess the question writers intentionally set a trap as this is one of the answer choices)
Probability of rain on Saturday OR Probability of rain on Sunday (given that it didn't rain on Saturday) OR Probability of rain on Monday (given that it didn't rain on Sunday) = 0.2 + (0.8 x 0.2) + (0.8 x 0.8 x 0.2) = 0.2 + 0.16 + (0.128 <- The actual answer!) =0.488 (C)
Re: On Saturday morning, Malachi will begin a camping vacation
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05 Apr 2017, 22:16
udaymathapati wrote:
On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday?
A. 0.008 B. 0.128 C. 0.488 D. 0.512 E. 0.640
I would've gotten this question wrong anyway, but my understanding of "on the following Monday is:
Re: On Saturday morning, Malachi will begin a camping vacation
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24 Aug 2017, 12:12
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Top Contributor
udaymathapati wrote:
On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday?
A. 0.008 B. 0.128 C. 0.488 D. 0.512 E. 0.640
NOTE: if P(rain on a certain day) = 0.2, then we know that P(NO rain on a certain day) = 1 - 0.2 = 0.8
For probability questions, I always ask, "What needs to happen for the desired event to occur?"
For this question P(come home Monday night) = P(no rain on Saturday AND no rain on Sunday AND rain on Monday)
At this point, we can apply what we know about AND probabilities. We get: P(come home Monday night) = P(no rain on Saturday) X P(no rain on Sunday) X P(rain on Monday) = (0.8) X (0.8) X (0.2) = 0.128
Re: On Saturday morning, Malachi will begin a camping vacation
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10 Jan 2018, 20:20
Hi All,
The standard approach to these types of probability questions is to determine the probability of each individual 'event', then multiply those probabilities together. With this question, there's a minor Number Property 'shortcut' at the end that can save you some time (and it helps if you're paying attention to how the answers are 'spaced out.'
Here, we're told: -the probability of rain occurring on any individual day = 0.2 -thus, the probability of rain NOT occurring on any individual day = 1 - 0.2 = 0.8
For Malachi to return at the end of the day on Monday, the following series of events must occur: (No rain on Saturday)(No rain on Sunday)(Rain on Monday).
The probability of that exact chain of events is: (.8)(.8)(.2)
At this point, you could just multiply those numbers together, but here's that math shortcut I referred to earlier: multiplying any positive number by a positive fraction (between 0 and 1) will result in a SMALLER number. Since we're multiplying 3 positive fractions together, the result WILL be less than 0.2.... Thus, the correct answer MUST be either A or B.
You probably already know that 2x2x2 = 8. IF.... you multiplied (.2)(.2)(.2), you would end up with .008 (re: Answer A) - but this is clearly SMALLER than the product that will actually occur (because two of those .2s are actually .8s), so the correct answer CANNOT be A.
Re: On Saturday morning, Malachi will begin a camping vacation
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20 May 2018, 04:44
GMATPrepNow wrote:
udaymathapati wrote:
On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday?
A. 0.008 B. 0.128 C. 0.488 D. 0.512 E. 0.640
NOTE: if P(rain on a certain day) = 0.2, then we know that P(NO rain on a certain day) = 1 - 0.2 = 0.8
For probability questions, I always ask, "What needs to happen for the desired event to occur?"
For this question P(come home Monday night) = P(no rain on Saturday AND no rain on Sunday AND rain on Monday)
At this point, we can apply what we know about AND probabilities. We get: P(come home Monday night) = P(no rain on Saturday) X P(no rain on Sunday) X P(rain on Monday) = (0.8) X (0.8) X (0.2) = 0.128
in the video above there is probability problem (what is the probabilty that the product of two picked numbers is positive from this set of numbers( -4, -3, -2, -1, 1, 2, 3 )
i solved it using combinatorics formula (see below), but i couldnt solve it using probability method/formula for some reason
# of total out comes is \(C^2_7=21\)
probability that two numbers are negatives is \(C^2_4=6\)
probability that two numbers are positive is \(C^2_3=3\)
total number of desired probabilities \(C^2_4=6\) + \(C^2_3=3\) = \(9\)
\(\frac{9}{21}\) = \(\frac{3}{7}\)
Now using probability method
now product of two numbers is positive in case two numbers are negative or two numbers are positive
case one: two numbers negative - > there are 4 negative numbers so probability \(\frac{1}{4}\)*\(\frac{1}{3}\)*\(\frac{1}{2}\)*\(1\) = \(\frac{1}{24}\)
case two: two numbers are positive --> there are 3 positive numbers so probabilityis \(\frac{1}{3}\)*\(\frac{1}{2}\)*\(1\) =\(\frac{1}{6}\)
\(\frac{1}{6} +\frac{1}{24} =\frac{5}{24}\)
now i need to find tital number of outcomes ...how to do it using probability formula without using combinatorics formula ?
Please find the method to do the problem if you used probability.
Out of the 7 numbers, we have 3 numbers that are positive and 4 that are negative. We need to find the number of cases where the product of any two random numbers is positive.
When the first number picked is negative and second number picked is positive: \(\frac{4}{7}*\frac{3}{6} = \frac{2}{7}\) When the first number picked is positive and second number picked is negative: \(\frac{3}{7}*\frac{4}{6} = \frac{2}{7}\)
Combined cases when the product can be negative is \(\frac{4}{7}(\frac{2}{7}+\frac{2}{7}\))
P(Product being positive) = 1 - P(Product being negative) Therefore, the cases when the product is positive is \(1 - \frac{4}{7} = \frac{3}{7}\)
Hope this helps you.
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