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Abhishek, good reply; beat me to it by 2 minutes! EDIT: But it does need to say 0.9 * 0.9 * 0.1. Shame on me for not reading closely enough.

Fun twist: If you know that Nora did, in fact, win on one of the first three days, what's the probability she won on day 3?
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ziyuenlau
Nora will enter a ticket lottery every day until she wins the lottery, after which she will no longer enter. If the probability that she wins the ticket lottery is \(0.1\) on each of the first three days, what is the probability that she wins on the third day?

(A) \(0.001\)
(B) \(0.009\)
(C) \(0.081\)
(D) \(0.729\)
(E) \(0.900\)


I don't see a correct answer. If the probability she wins on any given day of the first three days is 0.1 the "the probability she wins on the third day" would still be 0.1

Look carefully again at the highlighted part....

Nora entered Ticket Lottery coz she didn't win it on the first 2 days..

Probability of not winning lottery on the first 2 days is 0.9*0.9 = 0.081

Hence, correct answer must be (C) 0.081


You're missing a step though right?

Probability of not winning lottery on the first 2 days is 0.9*0.9=0.81 not 0.081 yet.

The probability of winning on the third day is 0.81*0.1 = 0.081 (Probability of making it to the third day x probability of winning on day 3).
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ziyuenlau
Nora will enter a ticket lottery every day until she wins the lottery, after which she will no longer enter. If the probability that she wins the ticket lottery is \(0.1\) on each of the first three days, what is the probability that she wins on the third day?

(A) \(0.001\)
(B) \(0.009\)
(C) \(0.081\)
(D) \(0.729\)
(E) \(0.900\)

We are given that Nora’s probability of winning the lottery on each of the first 3 days is 0.1; thus, the probability of her not winning on any day is 1 - 0.1 = 0.9. We must determine the probability of her winning on day 3, which means she does not win on day 1 or on day 2.

P(winning on day 3) = 0.9 x 0.9 x 0.1 = 0.081.

Answer: C
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Since this question gives you the probability of winning (1/10), it is implied that the probability of losing is (9/10)

The question basically asks: (9/10)*(9/10)*(1/10) = ?
-- 81/1,000 --> .081
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Nora will enter a ticket lottery every day until she wins the lottery, after which she will no longer enter. If the probability that she wins the ticket lottery is \(0.1\) on each of the first three days, what is the probability that she wins on the third day?

(A) \(0.001\)
(B) \(0.009\)
(C) \(0.081\)
(D) \(0.729\)
(E) \(0.900\)

Actually probability of Nora winning on any one single day is the same , be it the third day or second day or first day.

Consider the following :

Probability of winning on third day = 0.9*0.9*0.1 = .081
Probability of winning on second day = 0.9*0.1*0.9 = .081
Probability of winning on First day = 0.1*0.9*0.9 = .081

Hope this helps.
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hazelnut
Nora will enter a ticket lottery every day until she wins the lottery, after which she will no longer enter. If the probability that she wins the ticket lottery is \(0.1\) on each of the first three days, what is the probability that she wins on the third day?

(A) \(0.001\)
(B) \(0.009\)
(C) \(0.081\)
(D) \(0.729\)
(E) \(0.900\)

She wins on the third day, therefore she was lost in her first 2 days:

The probability that she will lost on each of the first 3 days: \(1 - 0.1 = 0.9\)
Probability that she wins on the third day: \(P(Lost in day 1) * P(Lost in day 2) * P(Win in day 3) = 0.9*0.9*0.1 = 0.081\) => Answer (C)
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hazelnut
Nora will enter a ticket lottery every day until she wins the lottery, after which she will no longer enter. If the probability that she wins the ticket lottery is \(0.1\) on each of the first three days, what is the probability that she wins on the third day?

(A) \(0.001\)
(B) \(0.009\)
(C) \(0.081\)
(D) \(0.729\)
(E) \(0.900\)

Actually probability of Nora winning on any one single day is the same , be it the third day or second day or first day.

Consider the following :

Probability of winning on third day = 0.9*0.9*0.1 = .081
Probability of winning on second day = 0.9*0.1*0.9 = .081
Probability of winning on First day = 0.1*0.9*0.9 = .081

Hope this helps.

Since she enters the lottery only until she wins do you still factor in the .9 for the other days?
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stne
hazelnut
Nora will enter a ticket lottery every day until she wins the lottery, after which she will no longer enter. If the probability that she wins the ticket lottery is \(0.1\) on each of the first three days, what is the probability that she wins on the third day?

(A) \(0.001\)
(B) \(0.009\)
(C) \(0.081\)
(D) \(0.729\)
(E) \(0.900\)

Actually probability of Nora winning on any one single day is the same , be it the third day or second day or first day.

Consider the following :

Probability of winning on third day = 0.9*0.9*0.1 = .081
Probability of winning on second day = 0.9*0.1*0.9 = .081
Probability of winning on First day = 0.1*0.9*0.9 = .081

Hope this helps.

Since she enters the lottery only until she wins do you still factor in the .9 for the other days?

He made a mistake as she only enters the next day if she didn't win yesterday. So if she wins on day 1, she won't enter in day 2 onward, or if she wins in day 2, she will only enter it in day 1 and 2 but not 3. But the point is still correct - the probability of her winning on any given day is the same because she only wins once (you won't have to take orders of her winning into consideration)
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hoang221

He made a mistake as she only enters the next day if she didn't win yesterday. So if she wins on day 1, she won't enter on day 2 onward, or if she wins on day 2, she will only enter it in day 1 and 2 but not 3. But the point is still correct - the probability of her winning on any given day is the same because she only wins once (you won't have to take orders of her winning into consideration)

That's what I thought but I let it get in my head. Although in this case is the probability of her winning on the first or second day, not 0? Given that she only plays until she wins and that the probability of her winning on days 2 and 3 are .1 doesn't that mean that she lost for sure on days one and 2...
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hoang221

He made a mistake as she only enters the next day if she didn't win yesterday. So if she wins on day 1, she won't enter on day 2 onward, or if she wins on day 2, she will only enter it in day 1 and 2 but not 3. But the point is still correct - the probability of her winning on any given day is the same because she only wins once (you won't have to take orders of her winning into consideration)

That's what I thought but I let it get in my head. Although in this case is the probability of her winning on the first or second day, not 0? Given that she only plays until she wins and that the probability of her winning on days 2 and 3 are .1 doesn't that mean that she lost for sure on days one and 2...

This is a common misconception about probability. Although we know for sure that she would lose on day 1 and 2 (according to this problem), it actually not the case. We "know" that she will lose on day 1 and 2 because we assumed so. If the problem stated otherwise, we will have different combination of 0.1 and 0.9, but those 2 numbers will stay the same. The overall probability that she will win or lose in a single day will be unchanged
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hoang221

This is a common misconception about probability. Although we know for sure that she would lose on day 1 and 2 (according to this problem), it actually not the case. We "know" that she will lose on day 1 and 2 because we assumed so. If the problem stated otherwise, we will have different combination of 0.1 and 0.9, but those 2 numbers will stay the same. The overall probability that she will win or lose in a single day will be unchanged

No, I'm with you on that, I was referring to the wording of the question, the probability that she wins the lottery on each of the first 3 days is 0.1 does not suggest anything about day 4 but semantically with the information, we are provided we are given that "she no longer enters when she wins", so the probability of her winning on a day following a win is 0 since she doesn't play on those days.
I followed the question, I'm just saying if the question was worded the same except the end said " what is the probability that she wins on the second day?" instead of "what is the probability that she wins on the third day?" without changing the wording before it to something like "If the probability of winning" instead of "the probability that she wins".
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hawesg
hoang221

This is a common misconception about probability. Although we know for sure that she would lose on day 1 and 2 (according to this problem), it actually not the case. We "know" that she will lose on day 1 and 2 because we assumed so. If the problem stated otherwise, we will have different combination of 0.1 and 0.9, but those 2 numbers will stay the same. The overall probability that she will win or lose in a single day will be unchanged

No, I'm with you on that, I was referring to the wording of the question, the probability that she wins the lottery on each of the first 3 days is 0.1 does not suggest anything about day 4 but semantically with the information, we are provided we are given that "she no longer enters when she wins", so the probability of her winning on a day following a win is 0 since she doesn't play on those days.
I followed the question, I'm just saying if the question was worded the same except the end said " what is the probability that she wins on the second day?" instead of "what is the probability that she wins on the third day?" without changing the wording before it to something like "If the probability of winning" instead of "the probability that she wins".

Okay i think i understand what you meant by now
So what you imply is that the wording can throw us off, because if they ask "what is the probability that she wins on the third day?", the answer can be 0.1, in fact it will be 0.1 no matter what day she won. In this case, we have to rely on the answer choices (clearly there are no 0.1). I think this kind of problem will never happen on GMAT, as we will be required to do at least some calculation in order to reach the answer.
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hoang221
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hoang221

This is a common misconception about probability. Although we know for sure that she would lose on day 1 and 2 (according to this problem), it actually not the case. We "know" that she will lose on day 1 and 2 because we assumed so. If the problem stated otherwise, we will have different combination of 0.1 and 0.9, but those 2 numbers will stay the same. The overall probability that she will win or lose in a single day will be unchanged

No, I'm with you on that, I was referring to the wording of the question, the probability that she wins the lottery on each of the first 3 days is 0.1 does not suggest anything about day 4 but semantically with the information, we are provided we are given that "she no longer enters when she wins", so the probability of her winning on a day following a win is 0 since she doesn't play on those days.
I followed the question, I'm just saying if the question was worded the same except the end said " what is the probability that she wins on the second day?" instead of "what is the probability that she wins on the third day?" without changing the wording before it to something like "If the probability of winning" instead of "the probability that she wins".

Okay i think i understand what you meant by now
So what you imply is that the wording can throw us off, because if they ask "what is the probability that she wins on the third day?", the answer can be 0.1, in fact it will be 0.1 no matter what day she won. In this case, we have to rely on the answer choices (clearly there are no 0.1). I think this kind of problem will never happen on GMAT, as we will be required to do at least some calculation in order to reach the answer.

Agreed, and I think the question is perfectly clear the way it is. I was more trying to make sure my reasoning was correct in response to the comment that actually the odds would be the same if the question had asked what the probability of her winning on the first or second day was (if the rest of the question remained the same).

Cheers.
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