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Intern Joined: 09 Nov 2017
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The probability of rain in Greg's town... [#permalink] 00:00

Question Stats: 73% (00:50) correct 26% (00:45) wrong based on 15 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

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.
Edited by Carcass GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Re: The probability of rain in Greg's town... [#permalink]
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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

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

 1 Rain Quiz 2 Rain No Quiz 3 Rain No Quiz 4 Rain No Quiz 5 Rain No Quiz 6 Rain Quiz 7 Rain No Quiz 8 Rain No Quiz 9 Rain No Quiz 10 Rain No Quiz 11 Rain Quiz 12 Rain No Quiz 13 Rain No Quiz 14 Rain No Quiz 15 Rain No Quiz 16 No Rain Quiz 17 No Rain No Quiz 18 No Rain No Quiz 19 No Rain No Quiz 20 No Rain No Quiz 21 No Rain Quiz 22 No Rain No Quiz 23 No Rain No Quiz 24 No Rain No Quiz 25 No Rain No Quiz 26 No Rain Quiz 27 No Rain No Quiz 28 No Rain No Quiz 29 No Rain No Quiz 30 No Rain No Quiz 31 No Rain Quiz 32 No Rain No Quiz 33 No Rain No Quiz 34 No Rain No Quiz 35 No Rain No Quiz 36 No Rain Quiz 37 No Rain No Quiz 38 No Rain No Quiz 39 No Rain No Quiz 40 No Rain No Quiz 41 No Rain Quiz 42 No Rain No Quiz 43 No Rain No Quiz 44 No Rain No Quiz 45 No Rain No Quiz 46 No Rain Quiz 47 No Rain No Quiz 48 No Rain No Quiz 49 No Rain No Quiz 50 No Rain No 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?) Manager Joined: 22 Feb 2018
Posts: 158
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Re: The probability of rain in Greg's town... [#permalink]
1
KUDOS
P(rain in town on Tuesday) = 0.3 = P(r)
P(pop quiz on Tuesday) = 0.2 = P(Q)
Events are independent, So:
P(not raining in town on Tuesday) = 1 - 0.3 = 0.7
P(not pop quiz on Tuesday) = 1 - 0.2 = 0.8

A: The probability that either or both events occur =
P(r and not(Q)) + P(Q and not(r)) + P(Q)*P(r) =
P(r)*P(not(Q)) + P(Q)*P(not(r)) + P(Q)*P(r) =
0.3 * 0.8 + 0.2*0.7 + 0.3*0.2 = 0.44

B: The probability that neither event occurs =
P(not(Q) and not(r)) = P(not(Q)) * P(not(r)) =
0.7 * 0.8 = 0.56

So B is bigger than A.
**We write P(a and b) as P(a)*P(b) because a and b are independent events, otherwise we could’t write them in this way.
_________________ Re: The probability of rain in Greg's town...   [#permalink] 27 Dec 2018, 16:01
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