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The probability of rain in Greg's town...
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Updated on: 10 Nov 2017, 01:27
Updated on: 10 Nov 2017, 01:27
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69% (01:17) correct
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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.
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.
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.
Show: ::
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
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
Re: The probability of rain in Greg's town...
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27 Dec 2018, 16:01
27 Dec 2018, 16:01
5
1
Answer: B
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.
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.
General Discussion
Re: The probability of rain in Greg's town...
[#permalink]
09 Nov 2017, 18:19
09 Nov 2017, 18:19
2
Expert Reply
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
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
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
