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23 Jun 2026, 12:42
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Difficulty:
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Question Stats:
38% (02:58) correct
62%
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wrong
based on 37
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Paco and six friends each select one date at random from the possible dates for a farewell lunch. If the probability that exactly one friend selects the same date as Paco is one-third the probability that no friend selects the same date as Paco, how many possible dates for the farewell lunch are there?
A. 10
B. 18
C. 19
D. 21
E. 22
A. 10
B. 18
C. 19
D. 21
E. 22
23 Jun 2026, 16:39
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Good question! On complicated probability questions, it's helpful to break them into their component parts in order to make them more manageable.
Here, you're comparing two different probabilities:
Scenario 1. The probability that exactly one friend picks the same date as Paco
Scenario 2. The probability that no friend picks the same date as Paco
We know that the first is one-third of the second, but it will be much easier to calculate them separately and then put them together.
Also, since we don't know how many dates there are, we can't calculate the probability from what's given in the prompt. Instead, we need to work "backwards", which means that we should start with the answer choices. Let's start with choice A, since it's a nice round number (10).
Scenario 1. One friend picks the same date
Think it through. If there are 10 dates and Paco picked one of them, any friend has a 1 in 10 chance of picking the same one. Then the other 5 friends each have a 9 in 10 chance of picking a different date. For all of these things to happen, we multiply:
\((\frac{1}{10})(\frac{9}{10})^5\)
This is the same probability for each of the 6 friends, so multiply by 6:
\(6(\frac{1}{10})(\frac{9}{10})^5\)
Scenario 2. No friend picks the same date
If there are 10 dates and Paco picked one of them, his friends each have a 9 in 10 chance of picking a different date. Since there are 6 friends, we have:
\((\frac{9}{10})^6\)
Now compare the two results. They each have \(9^5\) and \((\frac{1}{10})^6\), with 6 "left over" for scenario 1 and 9 "left over" for scenario 2. 6 isn't one-third of 9, so this isn't correct.
What should you try next? Note that the "6" from scenario 1 is not going to change (the number of friends will stay at 6). But the "9" from scenario 2 will change: it represents the dates that are NOT Paco's, so it must be 1 less than the total number of dates. This implies that we should try 19 dates next, since 1 less than 19 is 18, and 6 is one-third of 18.
To confirm, let's run it through with 19 dates:
Scenario 1: If there are 19 dates and Paco picked one of them, any friend has a 1 in 19 chance of picking the same one. Then the other 5 friends each have a 18 in 19 chance of picking a different date:
\(6(\frac{1}{19})(\frac{18}{19})^5\)
Scenario 2: If there are 19 dates and Paco picked one of them, his friends each have a 18 in 19 chance of picking a different date. Since there are 6 friends, we have:
\((\frac{18}{19})^6\)
Again, compare the results. They each have \(18^5\) and \((\frac{1}{19})^6\), leaving 6 for scenario 1 and 18 for scenario 2. 6 is one-third of 18, so answer choice C is correct.
Remember, when you have a probability question with multiple parts, it's generally best to break it into its component scenarios before comparing. And in a problem like this where you're asked to work backwards to determine the number of possibilities, backsolving is the preferred strategy.
Here, you're comparing two different probabilities:
Scenario 1. The probability that exactly one friend picks the same date as Paco
Scenario 2. The probability that no friend picks the same date as Paco
We know that the first is one-third of the second, but it will be much easier to calculate them separately and then put them together.
Also, since we don't know how many dates there are, we can't calculate the probability from what's given in the prompt. Instead, we need to work "backwards", which means that we should start with the answer choices. Let's start with choice A, since it's a nice round number (10).
Scenario 1. One friend picks the same date
Think it through. If there are 10 dates and Paco picked one of them, any friend has a 1 in 10 chance of picking the same one. Then the other 5 friends each have a 9 in 10 chance of picking a different date. For all of these things to happen, we multiply:
\((\frac{1}{10})(\frac{9}{10})^5\)
This is the same probability for each of the 6 friends, so multiply by 6:
\(6(\frac{1}{10})(\frac{9}{10})^5\)
Scenario 2. No friend picks the same date
If there are 10 dates and Paco picked one of them, his friends each have a 9 in 10 chance of picking a different date. Since there are 6 friends, we have:
\((\frac{9}{10})^6\)
Now compare the two results. They each have \(9^5\) and \((\frac{1}{10})^6\), with 6 "left over" for scenario 1 and 9 "left over" for scenario 2. 6 isn't one-third of 9, so this isn't correct.
What should you try next? Note that the "6" from scenario 1 is not going to change (the number of friends will stay at 6). But the "9" from scenario 2 will change: it represents the dates that are NOT Paco's, so it must be 1 less than the total number of dates. This implies that we should try 19 dates next, since 1 less than 19 is 18, and 6 is one-third of 18.
To confirm, let's run it through with 19 dates:
Scenario 1: If there are 19 dates and Paco picked one of them, any friend has a 1 in 19 chance of picking the same one. Then the other 5 friends each have a 18 in 19 chance of picking a different date:
\(6(\frac{1}{19})(\frac{18}{19})^5\)
Scenario 2: If there are 19 dates and Paco picked one of them, his friends each have a 18 in 19 chance of picking a different date. Since there are 6 friends, we have:
\((\frac{18}{19})^6\)
Again, compare the results. They each have \(18^5\) and \((\frac{1}{19})^6\), leaving 6 for scenario 1 and 18 for scenario 2. 6 is one-third of 18, so answer choice C is correct.
Remember, when you have a probability question with multiple parts, it's generally best to break it into its component scenarios before comparing. And in a problem like this where you're asked to work backwards to determine the number of possibilities, backsolving is the preferred strategy.
26 Jun 2026, 11:37
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Assuming there are n days (D1, D2...Dn)
P(F=Dx | Paco = Dx) = 1/n * ((n-1)/n)^5 ------ Eq 1 (5 friends shouldn't choose Dx)
You can have any of the 6 friends so 6C1*Eq 1 which is equal to one-third of ((n-1)/n)^6 ---- Eq 2
Solving for Eq1 == Eq2
18n = n^2 -n
thus n = 19
P(F=Dx | Paco = Dx) = 1/n * ((n-1)/n)^5 ------ Eq 1 (5 friends shouldn't choose Dx)
You can have any of the 6 friends so 6C1*Eq 1 which is equal to one-third of ((n-1)/n)^6 ---- Eq 2
Solving for Eq1 == Eq2
18n = n^2 -n
thus n = 19
