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The probability of rain in Greg's town...

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The probability of rain in Greg's town... [#permalink] New post 09 Nov 2017, 13:55
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Question Stats:

87% (00:42) correct 12% (00:00) wrong based on 8 sessions
The probability of rain in Greg's town on Tuesday is 0.3. The probability that Greg's teacher will give him a pop quiz on Tuesday is 0.2. The events occur independently of each other.

Quantity A
Quantity B
The probability that either or both events occur
The probability that neither event occurs


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Could somebody PLEASE explain the reasoning in the answer. Many thanks.

[Reveal] Spoiler:
HELP

Answer: B

The problem indicated that the events occur independently of each other. Therefore in calculating Quantity do not just add both events. Adding 0.3+0.2 is incorrect because the probability that both events occur is counted twice. While Quantity A should include the probability that both events occur, make sure to count this probability only once, not twice. Since the probability that both events occur is 0.3(0.2) = 0.06 subtract this value from the "or" probability.
q
[Reveal] Spoiler: OA

Last edited by Carcass on 10 Nov 2017, 01:27, edited 1 time in total.
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Re: The probability of rain in Greg's town... [#permalink] New post 09 Nov 2017, 18:19
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Expert's post
Since the explanation given is a standard one. I will try to in much more detailed way!

Event of rain and event of quiz are independent of each other.

So we have the following combination

Rain + Quiz
Rain + No Quiz
No Rain + Quiz
No Rain + No Quiz

Lets say we have 50 day. Then we can write





1RainQuiz
2RainNo Quiz
3RainNo Quiz
4RainNo Quiz
5RainNo Quiz
6RainQuiz
7RainNo Quiz
8RainNo Quiz
9RainNo Quiz
10RainNo Quiz
11RainQuiz
12RainNo Quiz
13RainNo Quiz
14RainNo Quiz
15RainNo Quiz
16No RainQuiz
17No RainNo Quiz
18No RainNo Quiz
19No RainNo Quiz
20No RainNo Quiz
21No RainQuiz
22No RainNo Quiz
23No RainNo Quiz
24No RainNo Quiz
25No RainNo Quiz
26No RainQuiz
27No RainNo Quiz
28No RainNo Quiz
29No RainNo Quiz
30No RainNo Quiz
31No RainQuiz
32No RainNo Quiz
33No RainNo Quiz
34No RainNo Quiz
35No RainNo Quiz
36No RainQuiz
37No RainNo Quiz
38No RainNo Quiz
39No RainNo Quiz
40No RainNo Quiz
41No RainQuiz
42No RainNo Quiz
43No RainNo Quiz
44No RainNo Quiz
45No RainNo Quiz
46No RainQuiz
47No RainNo Quiz
48No RainNo Quiz
49No RainNo Quiz
50No RainNo Quiz


Now if you count the number of days with rain and quiz it is 3. Day 1 Day 6 and day 11.

Now How did we take 50 days.

It is given that probability of rain is 0.3 or \(\frac{3}{10}\).
probability of quiz is 0.2 or \(\frac{1}{5}\).

50 is the product of denominator 10 and 5. You need 50 days to explain every possible combination of rain and quiz.

How do you make the table?
Rain occurs 15 times in 50 days and no rain is 35. Write all the 15 rain days on top and 35 no rain days after that in a coulmn.
Now Quiz occurs once in every 5 days. Write Quiz on top followed by 4 no quiz. Repeat the pattern till you fill all 50 days.

Now this way you will always find the answer, but it is impossible to write such a huge table. Also is there are 3 or 4 events then the number of samples neefed would be much larger.

Probability of Event 1 and Event 2 happening = P(1)*P(2) where 1 and 2 are independent.

Going back to our 50 days example:
How many days is it going to rain out of 50 days? = \(50*\frac{3}{10}\)= 15
How many days do we have tests out of 15 days? =\(15*\frac{1}{5}\)= 3
How many days do we have rain and test out of 50?= 3

Probability of rain and test = \(\frac{3}{50}\)= 0.06.

Did you spot the similarity when we mutiplied the first probability to 50 days we got the first 15 rows of out table


Similarly Probability of no rain and no test = \((1-0.3)*(1-0.2)=0.56\).

Again if you count from the table you get 28 days out of 50 with no rain and no test.

So quanity B is greater!

Hope you can vizualize the problem now!

Cheers
_________________

Sandy
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Re: The probability of rain in Greg's town... [#permalink] New post 10 Nov 2017, 04:13
sandy wrote:
Since the explanation given is a standard one. I will try to in much more detailed way!

Event of rain and event of quiz are independent of each other.

So we have the following combination

Rain + Quiz
Rain + No Quiz
No Rain + Quiz
No Rain + No Quiz

Lets say we have 50 day. Then we can write





1RainQuiz
2RainNo Quiz
3RainNo Quiz
4RainNo Quiz
5RainNo Quiz
6RainQuiz
7RainNo Quiz
8RainNo Quiz
9RainNo Quiz
10RainNo Quiz
11RainQuiz
12RainNo Quiz
13RainNo Quiz
14RainNo Quiz
15RainNo Quiz
16No RainQuiz
17No RainNo Quiz
18No RainNo Quiz
19No RainNo Quiz
20No RainNo Quiz
21No RainQuiz
22No RainNo Quiz
23No RainNo Quiz
24No RainNo Quiz
25No RainNo Quiz
26No RainQuiz
27No RainNo Quiz
28No RainNo Quiz
29No RainNo Quiz
30No RainNo Quiz
31No RainQuiz
32No RainNo Quiz
33No RainNo Quiz
34No RainNo Quiz
35No RainNo Quiz
36No RainQuiz
37No RainNo Quiz
38No RainNo Quiz
39No RainNo Quiz
40No RainNo Quiz
41No RainQuiz
42No RainNo Quiz
43No RainNo Quiz
44No RainNo Quiz
45No RainNo Quiz
46No RainQuiz
47No RainNo Quiz
48No RainNo Quiz
49No RainNo Quiz
50No RainNo Quiz


Now if you count the number of days with rain and quiz it is 3. Day 1 Day 6 and day 11.

Now How did we take 50 days.

It is given that probability of rain is 0.3 or \(\frac{3}{10}\).
probability of quiz is 0.2 or \(\frac{1}{5}\).

50 is the product of denominator 10 and 5. You need 50 days to explain every possible combination of rain and quiz.

How do you make the table?
Rain occurs 15 times in 50 days and no rain is 35. Write all the 15 rain days on top and 35 no rain days after that in a coulmn.
Now Quiz occurs once in every 5 days. Write Quiz on top followed by 4 no quiz. Repeat the pattern till you fill all 50 days.

Now this way you will always find the answer, but it is impossible to write such a huge table. Also is there are 3 or 4 events then the number of samples neefed would be much larger.

Probability of Event 1 and Event 2 happening = P(1)*P(2) where 1 and 2 are independent.

Going back to our 50 days example:
How many days is it going to rain out of 50 days? = \(50*\frac{3}{10}\)= 15
How many days do we have tests out of 15 days? =\(15*\frac{1}{5}\)= 3
How many days do we have rain and test out of 50?= 3

Probability of rain and test = \(\frac{3}{50}\)= 0.06.

Did you spot the similarity when we mutiplied the first probability to 50 days we got the first 15 rows of out table


Similarly Probability of no rain and no test = \((1-0.3)*(1-0.2)=0.56\).

Again if you count from the table you get 28 days out of 50 with no rain and no test.

So quanity B is greater!

Hope you can vizualize the problem now!

Cheers


Thank you. That is helpful.

I think my difficulty is coming from subtracting the possibility of them both happening as I don't understand the concept of doing this when it asks for the probability of both.
If this was an OR problem i.e. probability of rain OR test, then I could understand subtracting the probability of both - but in this question I do not?

(If you do remove the probability of both, because its been counted "twice" - as the explanation suggests - then surely in OR problems you would have to remove the probability of both x 2 so that it isn't counted at all?)
Re: The probability of rain in Greg's town...   [#permalink] 10 Nov 2017, 04:13
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