langsend
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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.

Which is greater?

1. The probability that either or both events occur

2. The probability that neither event occurs

Explanation below but i don't understand

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Answer 1. 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.

I don't understand the subtraction of the probability both occur? If the question had asked under 1: the probability either event occurs I would have subtracted the possibility of both, but I don't understand how in this scenario its been counted twice? Thanks.
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Zacken
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Are you aware of P(A U B) = P(A) + P(B) - P(A n B)?
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langsend
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(Original post by Zacken)
Are you aware of P(A U B) = P(A) + P(B) - P(A n B)?

Sorry it has been a very long time since I have done maths.

I understand the concept of them intersecting, and so you have to remove the overlap. What is confusing me is the concept of if quantity 1 was defined as the probability of either event occurring (not including both) then would I remove 2 x P(A n B)
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RogerOxon
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(Original post by langsend)
I don't understand the subtraction of the probability both occur? If the question had asked under 1: the probability either event occurs I would have subtracted the possibility of both, but I don't understand how in this scenario its been counted twice?
You know the probability of A is 0.3 and B is 0.2, and that they are independent. You have a number of possible outcomes:

1. A only
2. B only
3. A & B
4. Neither A nor B

P(1)=0.3 * (1-0.2) =0.24
P(2)=(1-0.3) * 0.2 =0.14
P(3)=0.3 * 0,2 =0.06
P(4)=(1-0.3)*(1-0.2)=0.56

Adding them, you will see that we get 1.0, as we've covered all the possibilities. If you add the 0.3 and 0.2, you get 0.5, but the probability is actually 0.24+0.14+0.06=0.44, as the 0.5 counts the probability of both occurring twice - once in the 0.3 and once in the 0.2. Think about a 0.2 and 0.3 overlapped by 0.06.
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langsend
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(Original post by RogerOxon)
You know the probability of A is 0.3 and B is 0.2, and that they are independent. You have a number of possible outcomes:

1. A only
2. B only
3. A & B
4. Neither A nor B

P(1)=0.3 * (1-0.2) =0.24
P(2)=(1-0.3) * 0.2 =0.14
P(3)=0.3 * 0,2 =0.06
P(4)=(1-0.3)*(1-0.2)=0.56

Adding them, you will see that we get 1.0, as we've covered all the possibilities. If you add the 0.3 and 0.2, you get 0.5, but the probability is actually 0.24+0.14+0.06=0.44, as the 0.5 counts the probability of both occurring twice - once in the 0.3 and once in the 0.2. Think about a 0.2 and 0.3 overlapped by 0.06.
Thank you very much and apologies this is at such a basic level. I think my confusion has stemmed from me conceptually thinking that the individual probabilities added includes the possibility of them both occurring?

For example: in a problem where a die is rolled and a coin is flipped, if you are asked for probability that either the die will come up 2 or the coin will land on heads you would add the probability of either event and then minus the probability of both (but in this question it specifically asks you to exclude both and not include it like the one above.)
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RogerOxon
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(Original post by langsend)
For example: in a problem where a die is rolled and a coin is flipped, if you are asked for probability that either the die will come up 2 or the coin will land on heads you would add the probability of either event and then minus the probability of both (but in this question it specifically asks you to exclude both and not include it like the one above.)
Isn't that what you need to do here, i.e. 0.2 + 0.3 - 0.2*0.3 = 0.44?

You're essentially overlapping 0.2 and 0.3 by 0.06 in the probability range, so you need to remove 0.06 to get the correct probability or either / both.
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RogerOxon
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(Original post by langsend)
Thank you very much and apologies this is at such a basic level. I think my confusion has stemmed from me conceptually thinking that the individual probabilities added includes the possibility of them both occurring?
It includes that twice. Draw line 0.2 units long, and overlap it with another of 0.3 units long, by 0.06 units. What is the length of the line?
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langsend
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(Original post by RogerOxon)
Isn't that what you need to do here, i.e. 0.2 + 0.3 - 0.2*0.3 = 0.44?

You're essentially overlapping 0.2 and 0.3 by 0.06 in the probability range, so you need to remove 0.06 to get the correct probability or either / both.
Sorry you're literally having to take me through every step here.

In the coin/die problem it specifically asks you to exclude the possibility of both, whereas the original problem asks you to include this. If you have to remove the probability of both (as its counted twice) from the problem which wants you to include the probability of both, then why do you not need to remove the probability of both twice from the one which asks for the probability of either, but not both?
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RogerOxon
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(Original post by langsend)
Sorry you're literally having to take me through every step here.
I never did like probability, so it's good to have to think about it a bit.

In the coin/die problem it specifically asks you to exclude the possibility of both, whereas the original problem asks you to include this.
Where did it say that? I'm not seeing it.
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langsend
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(Original post by RogerOxon)
I never did like probability, so it's good to have to think about it a bit.


Where did it say that? I'm not seeing it.
In the rain and test problem: chance of rain AND/OR test: 3/10 + 2/10 = 5/10. But apparently this includes chance of both counted twice (im not sure how), so remove one of them 3/10 x 2/10 = 0.06. 0.5 - 0.06 = 0.44.

In the coin/die problem: chance of a heads OR a 2. 1/2 + 1/6 = 4/6. But need to remove chance of both (4/6 - (1/2 x 1/6)). But from reasoning above, if you add both the probabilities, this includes the chance of both x 2, so wouldnt you need to remove 2 x the chance of both?
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RogerOxon
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(Original post by langsend)
In the rain and test problem: chance of rain AND/OR test: 3/10 + 2/10 = 5/10. But apparently this includes chance of both counted twice (im not sure how), so remove one of them 3/10 x 2/10 = 0.06. 0.5 - 0.06 = 0.44.
The probabilities overlap. Think of two coin tosses - what's the probability of a heads on the first throw and / or on the second one? If you add, then you'd get 1.0 - clearly wrong. The correct answer is 0.75, as the 0.5 and 0.5 overlap by 0.25. We can only count that 0.25 once, getting (0.5-0.25) + 0.25 + (0.5-0.25) = 0.75.

In the coin/die problem: chance of a heads OR a 2. 1/2 + 1/6 = 4/6. But need to remove chance of both (4/6 - (1/2 x 1/6)). But from reasoning above, if you add both the probabilities, this includes the chance of both x 2, so wouldnt you need to remove 2 x the chance of both?
You've removed (once) the probability of both in each of those examples.

You only remove the chance of both once, because you still want it to be included (once) in the final answer.
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langsend
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(Original post by RogerOxon)
The probabilities overlap. Think of two coin tosses - what's the probability of a heads on the first throw and / or on the second one? If you add, then you'd get 1.0 - clearly wrong. The correct answer is 0.75, as the 0.5 and 0.5 overlap by 0.25. We can only count that 0.25 once, getting (0.5-0.25) + 0.25 + (0.5-0.25) = 0.75.


You've removed (once) the probability of both in each of those examples.

You only remove the chance of both once, because you still want it to be included (once) in the final answer.
Thank you so much.

I think maybe in the coin/die one, I was thinking it was the chance of either but NOT the chance of both happening, so you wouldn't want both included in the answer at all. My head hurts. I'm so sorry for dragging you through this with me haha
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