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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 13 May 2022, 05:15
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yvonne randomly picks an integer from x+1 to 7, inclusive. What is the probability that Yvonne picks the number 3?

(A) \(\frac{11}{180}\)

(B) \(\frac{13}{180}\)

(C) \(\frac{5}{36}\)

(D) \(\frac{1}{20}\)

(E) \(\frac{1}{5}\)




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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 13 May 2022, 07:04
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There are six options for X, each of which has a probability of 1/6. Let's explore each of the options for X.
If X=1, there are six options for Y (2, 3, 4, 5, 6, or, 7), each of which has a probability of 1/6. So the probability of X=1 and Y=3 is 1/6*1/6 = 1/36.
If X=2, there are five options for Y (3, 4, 5, 6, or, 7), each of which has a probability of 1/5. So the probability of X=2 and Y=3 is 1/6*1/5 = 1/30.
If X>2, 3 won't be available as an option for Y, so the probability of X={3,4,5,6} and Y=3 is zero.

(X=1, Y=3) and (X=2, Y=3) are the only two ways to "win" and we know each of their probabilities. Let's add them together to find the total probability of winning.

1/36 + 1/30 = 5/180 + 6/180 = 11/180

Answer choice A.
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 18 May 2022, 22:17
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There are six options for X, each of which has a probability of 1/6. Let's explore each of the options for X.
If X=1, there are six options for Y (2, 3, 4, 5, 6, or, 7), each of which has a probability of 1/6. So the probability of X=1 and Y=3 is 1/6*1/6 = 1/36.
If X=2, there are five options for Y (3, 4, 5, 6, or, 7), each of which has a probability of 1/5. So the probability of X=2 and Y=3 is 1/6*1/5 = 1/30.
If X>2, 3 won't be available as an option for Y, so the probability of X={3,4,5,6} and Y=3 is zero.

(X=1, Y=3) and (X=2, Y=3) are the only two ways to "win" and we know each of their probabilities. Let's add them together to find the total probability of winning.

1/36 + 1/30 = 5/180 + 6/180 = 11/180

Answer choice A.
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I approached this question in a different way and could not get the answer mentioned in the options.

Total num of ways in which two integers can be selected:
If 1st person chooses 1, 2nd can choose out of 6 integers, if he chooses 2, then 2nd can choose amongst 5 and so on..
So total number of ways= (6+5+4+3+2+1)=21
Out of these outcomes only two are the desired outcomes.(1,3) and (2,3).
So probability of getting 3 is 2/21.
Where am i going wrong? Please help.

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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 18 May 2022, 23:49
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GOT THE ANSWER EXPLAINED BUT SOMEHOW CANT RAP MY HEAD AROUND THE Fact you need to take the X probability and mutiply. I mean its not required as per question asked!!
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 19 May 2022, 06:16
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There are six options for X, each of which has a probability of 1/6. Let's explore each of the options for X.
If X=1, there are six options for Y (2, 3, 4, 5, 6, or, 7), each of which has a probability of 1/6. So the probability of X=1 and Y=3 is 1/6*1/6 = 1/36.
If X=2, there are five options for Y (3, 4, 5, 6, or, 7), each of which has a probability of 1/5. So the probability of X=2 and Y=3 is 1/6*1/5 = 1/30.
If X>2, 3 won't be available as an option for Y, so the probability of X={3,4,5,6} and Y=3 is zero.

(X=1, Y=3) and (X=2, Y=3) are the only two ways to "win" and we know each of their probabilities. Let's add them together to find the total probability of winning.

1/36 + 1/30 = 5/180 + 6/180 = 11/180

Answer choice A.


I approached this question in a different way and could not get the answer mentioned in the options.

Total num of ways in which two integers can be selected:
If 1st person chooses 1, 2nd can choose out of 6 integers, if he chooses 2, then 2nd can choose amongst 5 and so on..
So total number of ways= (6+5+4+3+2+1)=21
Out of these outcomes only two are the desired outcomes.(1,3) and (2,3).
So probability of getting 3 is 2/21.
Where am i going wrong? Please help.

Posted from my mobile device
Show more

You are correct that there are 21 possible outcomes, but they are not equally likely. We have a CONDITIONAL PROBABILITY issue. Let's look at an analogous situation.

If I work from home, I always eat soup for lunch. If I go to the office, I choose randomly from among six options (sandwich, salad, hamburger, chicken, steak, and pasta), each with equal probability, at the restaurant next door. I work from home half the time. What percentage of the time do I eat soup for lunch? By your methodology, there are a total of seven options and soup is one of the options, so 1/7? Nope.

It should be clear that I have soup 1/2 the days.

You can only determine the probability of winning by doing

(count the ways to win) and divide by (count the total number of ways)

when all the ways have an equal probability. In my example, as well as in the question, the conditional probability component means that you need to calculate the probability of each individual way to win and then add up each of those probabilities.

Make sense?
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 19 May 2022, 06:45
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There are six options for X, each of which has a probability of 1/6. Let's explore each of the options for X.
If X=1, there are six options for Y (2, 3, 4, 5, 6, or, 7), each of which has a probability of 1/6. So the probability of X=1 and Y=3 is 1/6*1/6 = 1/36.
If X=2, there are five options for Y (3, 4, 5, 6, or, 7), each of which has a probability of 1/5. So the probability of X=2 and Y=3 is 1/6*1/5 = 1/30.
If X>2, 3 won't be available as an option for Y, so the probability of X={3,4,5,6} and Y=3 is zero.

(X=1, Y=3) and (X=2, Y=3) are the only two ways to "win" and we know each of their probabilities. Let's add them together to find the total probability of winning.

1/36 + 1/30 = 5/180 + 6/180 = 11/180

Answer choice A.


I approached this question in a different way and could not get the answer mentioned in the options.

Total num of ways in which two integers can be selected:
If 1st person chooses 1, 2nd can choose out of 6 integers, if he chooses 2, then 2nd can choose amongst 5 and so on..
So total number of ways= (6+5+4+3+2+1)=21
Out of these outcomes only two are the desired outcomes.(1,3) and (2,3).
So probability of getting 3 is 2/21.
Where am i going wrong? Please help.

Posted from my mobile device

You are correct that there are 21 possible outcomes, but they are not equally likely. We have a CONDITIONAL PROBABILITY issue. Let's look at an analogous situation.

If I work from home, I always eat soup for lunch. If I go to the office, I choose randomly from among six options (sandwich, salad, hamburger, chicken, steak, and pasta), each with equal probability, at the restaurant next door. I work from home half the time. What percentage of the time do I eat soup for lunch? By your methodology, there are a total of seven options and soup is one of the options, so 1/7? Nope.

It should be clear that I have soup 1/2 the days.

You can only determine the probability of winning by doing

(count the ways to win) and divide by (count the total number of ways)

when all the ways have an equal probability. In my example, as well as in the question, the conditional probability component means that you need to calculate the probability of each individual way to win and then add up each of those probabilities.

Make sense?
Show more

Yes it makes sense.

So in the above question, the outcome 7 is more likely as compared to others because it is in all sets of possible values and hence we cannot assume that each outcome have equal probability (1/21).
So in all condional probability cases we need to take care of that. Am i correct?

Thanks a lot.
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 19 May 2022, 07:13
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You are correct that there are 21 possible outcomes, but they are not equally likely. We have a CONDITIONAL PROBABILITY issue.
Show more

Just in case this is confusing, this is not a "conditional probability" situation, the way that phrase is used in math. A conditional probability situation is one where we know in advance that a certain condition is true about the result. So a standard conditional probability question would read something like "someone rolls a die numbered from 1 through 6. If the person rolled a prime number, what is the probability the roll was an odd number?" The condition is "the person rolled a prime number". You could make the question in this thread into a conditional probability question by asking "If Yvonne picks the number 3, what is the probability that Xavier picked the number 1?" Then "Yvonne picks 3" is the condition, and we're no longer considering every possible outcome, only the ones in which the condition is true.

As written, though, the question in this thread is just one where the individual outcomes are not equally likely. It's like a question that asks "The probability of rain is 0.8 and the probability of sun is 0.2, what is the probability it rains on both Saturday and Sunday?" Here we have four outcomes, Rain+Rain, Rain+Sun, Sun+Rain or Sun+Sun, but the answer is not 1/4 because the four outcomes are not equally likely; rain happens far more often than sun.
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 19 May 2022, 08:06
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I approached this question in a different way and could not get the answer mentioned in the options.

Total num of ways in which two integers can be selected:
If 1st person chooses 1, 2nd can choose out of 6 integers, if he chooses 2, then 2nd can choose amongst 5 and so on..
So total number of ways= (6+5+4+3+2+1)=21
Out of these outcomes only two are the desired outcomes.(1,3) and (2,3).
So probability of getting 3 is 2/21.
Where am i going wrong? Please help.

Posted from my mobile device

You are correct that there are 21 possible outcomes, but they are not equally likely. We have a CONDITIONAL PROBABILITY issue. Let's look at an analogous situation.

If I work from home, I always eat soup for lunch. If I go to the office, I choose randomly from among six options (sandwich, salad, hamburger, chicken, steak, and pasta), each with equal probability, at the restaurant next door. I work from home half the time. What percentage of the time do I eat soup for lunch? By your methodology, there are a total of seven options and soup is one of the options, so 1/7? Nope.

It should be clear that I have soup 1/2 the days.

You can only determine the probability of winning by doing

(count the ways to win) and divide by (count the total number of ways)

when all the ways have an equal probability. In my example, as well as in the question, the conditional probability component means that you need to calculate the probability of each individual way to win and then add up each of those probabilities.

Make sense?

Yes it makes sense.

So in the above question, the outcome 7 is more likely as compared to others because it is in all sets of possible values and hence we cannot assume that each outcome have equal probability (1/21).
So in all condional probability cases we need to take care of that. Am i correct?

Thanks a lot.
Show more

Not quite.

Let's go back to my lunch analogy with a small change.

If I work from home, I always eat soup for lunch. If I go to the office, I choose randomly between two restaurants, and once I choose a restaurant, I choose randomly between their menu options. Restaurant A serves two options: sandwich or salad. Restaurant B serves two options: sandwich or hamburger. I work from home half the time. By your methodology, sandwich would be the most common lunch I eat? Nope. That's not possible since I eat soup whenever I work from home and I work from home half the time...so I eat soup half the time...can't eat sandwich the most.

Here are all the options for my lunch analogy with the probabilities calculated. See if you can figure out how to apply this to the original question.

Stay home and eat soup = 1/2 * 1 = 1/2
Restaurant A and eat sandwich = 1/4 * 1/2 = 1/8
Restaurant B and eat sandwich = 1/4 * 1/2 = 1/8
Restaurant A and eat salad = 1/4 * 1/2 = 1/8
Restaurant B and eat hamburger = 1/4 * 1/2 = 1/8

So the probability that I eat soup is 1/2, the probability that I eat a sandwich is 1/8 + 1/8 = 1/4, the probability that I eat salad is 1/8, and the probability that I eat hamburger is 1/8. Note that I can do a quick sanity check on my math by showing that the total probability equals 1 (1/2 + 1/4 + 1/8 + 1/8 = 1).
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 19 May 2022, 08:32
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You are correct that there are 21 possible outcomes, but they are not equally likely. We have a CONDITIONAL PROBABILITY issue.

Just in case this is confusing, this is not a "conditional probability" situation, the way that phrase is used in math. A conditional probability situation is one where we know in advance that a certain condition is true about the result. So a standard conditional probability question would read something like "someone rolls a die numbered from 1 through 6. If the person rolled a prime number, what is the probability the roll was an odd number?" The condition is "the person rolled a prime number". You could make the question in this thread into a conditional probability question by asking "If Yvonne picks the number 3, what is the probability that Xavier picked the number 1?" Then "Yvonne picks 3" is the condition, and we're no longer considering every possible outcome, only the ones in which the condition is true.

As written, though, the question in this thread is just one where the individual outcomes are not equally likely. It's like a question that asks "The probability of rain is 0.8 and the probability of sun is 0.2, what is the probability it rains on both Saturday and Sunday?" Here we have four outcomes, Rain+Rain, Rain+Sun, Sun+Rain or Sun+Sun, but the answer is not 1/4 because the four outcomes are not equally likely; rain happens far more often than sun.
Show more

What is the probability of Y=3 GIVEN THAT X=1?
What is the probability of Y=3 GIVEN THAT X=2?
What is the probability of Y=3 GIVEN THAT X=3?
What is the probability of Y=3 GIVEN THAT X=4?
What is the probability of Y=3 GIVEN THAT X=5?
What is the probability of Y=3 GIVEN THAT X=6?

Conditional probability is an ELEMENT of this problem.
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 19 May 2022, 11:00
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What is the probability of Y=3 GIVEN THAT X=5?
What is the probability of Y=3 GIVEN THAT X=6?

Conditional probability is an ELEMENT of this problem.
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I'm not trying to be pedantic, but if on that basis you're calling this a "conditional probability" question, then you'd call almost every two-step probability problem a conditional probability problem. If you take the question "What is the probability the sum of two dice rolls is even?" then if you break it down "What is the probability of an even sum given the first roll is odd?" and "What is the probability of an even sum given the first roll is even?" you have the same type of breakdown you've performed above. But no mathematician in the world would describe this dice problem as a conditional probability question except in the most vacuous sense.

I posted only because if a test taker is looking for an explanation of the principles used to solve this problem, that test taker will find them explained in a book long before that book covers conditional probability. You only need to understand the principles of conditional probability in questions where we learn a condition on the outcome, and not, as in this problem or as in my dice question, a condition on the first in a sequence of events.
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Xavier randomly picks an integer x from 1 to 6, inclusive, and then Yv 21 Jul 2025, 09:49
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