Paco and six friends each select one date at random from th

Question

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

Show Answer

Official Answer

C

A. 10
B. 18
C. 19
D. 21
E. 22

MarkGMATMentor · Answer

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 can work "backwards", which means that we 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 an effective strategy.

impatientdude · Answer

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

BongBideshini · Answer

Let n be the number of possible dates.
Paco chooses a date. Each of the 6 friends independently chooses one of the n dates.
Probability that no friend chooses Paco's date:
P(none) = ((n - 1)/n)^6
Probability that exactly one friend chooses Paco's date:

P(exactly one) = 6C1 × (1/n) × ((n - 1)/n)^5
Given:
P(exactly one) = (1/3) × P(none)
6 × (1/n) × ((n - 1)/n)^5 = (1/3) × ((n - 1)/n)^6
6/n = (1/3) × (n - 1)/n
18 = n - 1
n = 19
 Answer: C

ga821 · Answer

Hey,

Its Paco and  6  friends
So shouldn't it be  (\frac{9}{10})^6  
 6(\frac{1}{10})(\frac{9}{10})^6

Likewise for 19

SVaidyaraman · Answer

it's an equation in disguise. The "one-third" relationship between two probabilities becomes an algebraic equation, and almost everything cancels. The trick is to set up the ratio and not compute either probability fully.

Set up the scene. Paco picks a date; that fixes the "target." Each of his 6 friends independently picks from n dates. For each friend:
- P(matches Paco) = 1/n
- P(doesn't match) = (n−1)/n

The two probabilities.

No friend matches — all 6 miss:

$$P(	ext{none}) = \left(\frac{n-1}{n}ight)^6$$

Exactly one friend matches — choose which friend matches (6 ways), that one matches, the other 5 miss:

$$P(	ext{exactly one}) = \binom{6}{1}\left(\frac{1}{n}ight)\left(\frac{n-1}{n}ight)^5$$

Apply the condition: exactly-one is one-third of none.

$$\binom{6}{1}\left(\frac{1}{n}ight)\left(\frac{n-1}{n}ight)^5 = \frac{1}{3}\left(\frac{n-1}{n}ight)^6$$

Now watch the cancellation. Divide both sides by (n−1)5/n5 — it appears on both sides:

$$6 \cdot \frac{1}{n} = \frac{1}{3} \cdot \frac{n-1}{n}$$

Multiply both sides by n:

$$6 = \frac{n-1}{3}$$

$$n - 1 = 18 \quad\Rightarrow\quad n = 19$$

Answer: C. 19

The move that mattered: the two probabilities share almost all their factors — six of the seven terms are identical. Setting up the ratio lets that common bulk cancel, collapsing a messy expression into 6 = (n−1)/3. The transferable habit: when a problem relates two probabilities, write the ratio before computing either one — the shared structure usually cancels, and you're left with simple algebra.

egmat · Answer

Hi ga821,

You've actually got  Scenario 2 exactly right : with all  6  friends needing a  different  date, it's  (9/10)^6 . The fix is only in Scenario 1.

Here's the key idea you're missing:  "exactly one friend matches" means one specific friend lands on Paco's date -  and the other five do NOT . So you can't have all six friends carrying the "different date" factor.

Walk it through:

- Pick  which  friend matches - that's the  6  out front (or  6C1 ).
-  That  friend matches - probability  1/10 .
- The  remaining 5 friends  each pick a different date -  (9/10)^5 ,  not(9/10)^6 .

So Scenario 1 is  6 · (1/10) · (9/10)^5 . The exponent is  5 , because once one friend is "used up" as the match, only  5  friends are left to differ. Using  (9/10)^6  would be asking six friends to differ while also having one of them match - that's seven friends' worth of conditions on only six people.

The same logic gives the  (18/19)^5  in the  19 -date version.

Quick way to feel it - shrink the group.  Suppose Paco has just  2  friends and  n  dates. "Exactly one friend matches" means: choose the  1  matcher ( 2  ways), that friend matches ( 1/n ), and the  1  leftover friend differs ( (n-1)/n ). So it's  2 · (1/n) · ((n-1)/n)^1  - the exponent is  2 - 1 = 1 , one less than the total.

Same rule every time:  total friends minus the ones designated to match = the exponent on the "different" factor.

Answer: C

Paco and six friends each select one date at random from th

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