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GMAT Club World Cup 2026 (Day 5): The cockpit crew for a long-haul

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GamePine

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10 Jul 2026, 20:49

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There are 6 copilots and 2 will be picked for any crew so I calculated C(6,2) to get 15 possible pairs.
(6*5)/2 = 15.

3 pilots can take any pairing of copilots so that is 3*15 copilot combinations = 45 crews

Then we add potential Hugo crews which equals the number of pairs of copilots over 500 hours which equals C(3,2) = 3

45 +3 = 48 valid crews.

The probability Hugo is the pilots is 3 Hugo crews/ 48 valid crews = 3/48 = 1/16

The probability 1 copilot has less than 500 is C(3,1) copilot over 500 * C(3,1) copilot under 500 = 3*3 = 9 pairs and 3 pilots can be added to any of those pairs so 9*3 = 27 crews with exactly 1 copilot under 500 hours.

27/48 = 9/16.

There is a 1/16 chance of Hugo being the pilot and a 9/16 chance of exactly 1 copilot having less than 500 hours.

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10 Jul 2026, 21:32

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Total valid crews: the 3 non-Hugo pilots pair with any of C(6,2) = 15 copilot pairs (45 crews). and Hugo pairs with only the C(3,2) = 3 pairs from the qualified copilots (3 crews), so 48 total.

Hugo = 3/48 = 1/16

Exactly 1 under 500, Hugo's crews need both copilots >= 500, so they contribute 0. Non-Hugo pilots need only qualified and one unqualified copilot, 3 x 3 = 9 pairs each, so 3 x 9 = 27 crews. So, 27/48 = 9/16

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10 Jul 2026, 22:06

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Hugo is piloy
valid crews:
hugo 3C2=3
other piolts:3*6C2=45
Total=48

P(hugo)=3/48=1/16

exactly 1 copilot<500 hrs

only poss with non-Hugo pioltes= 3*(3*3)=27
p=27/48=9/16

Ans-1/16,9/16

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10 Jul 2026, 22:35

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Let 3 co pilot who can be with Hugo be H (having atleast 500 hours) and other be P.

Number of crews:

If Hugo is the pilot, 3 C 2 = 3
If other 3 are the pilots, = 3 * (6C2) = 45

Total crews = 45 + 3 = 48

1. Probality Hugo is the pilot:

= 3/48 = 1/16

2. Probablity exactly 1 copilot has less tham 500 hours:

1. No hugo crew cases,

2. For other 3 pilots, way sto choose the copilot = 3 *3C1 * 3C 1 = 27

Probability = 27/48 = 9/16

Answer:
Hugo is the pilot = 1/16
Exactly 1 copilot has less than 500 hours = 9/16

Bunuel

The cockpit crew for a long-haul flight must consist of 1 pilot and 2 copilots, selected from an available pool of 4 pilots and 6 copilots. One of the pilots, Hugo, is only certified to serve as pilot if each of the two selected copilots has at least 500 hours of flight experience, a qualification held by 3 of the 6 available copilots. The remaining pilots can fly with any copilot. A cockpit crew will be selected at random from all possible crew combinations that meet these conditions.

The two columns in the table refer to two separate cases, each based on the same random selection from all valid cockpit crews. In the table, for Hugo is the pilot, select the probability that Hugo is the pilot on the selected cockpit crew. For Exactly 1 copilot has less than 500 hours, select the probability that exactly 1 of the 2 copilots on the selected cockpit crew has less than 500 hours. Make only two selections, one in each column.

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10 Jul 2026, 23:15

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Bunuel

All possible crew combinations that meet the required conditions = 3C1*6C2 (Out of the 4 available pilots, there are only 3 pilots that can fly with any copilot) + 1C1*3C2 (The remaining pilot, Hugo, can only fly with 3 of the 6 available copilots)

Upon solving, we get total crew combinations as 48.

Probability that Hugo is the pilot on the selected cockpit crew = (1C1*3C2)/48 = 1/16

For the second column, we can choose any pilot out of the 3 pilots, any copilot out of the 3 copilots with less than 500 hours, and any copilot out of the 3 pilots with more than 500 hours. Therefore, probability that exactly 1 copilot on the selected cockpit crew has less than 500 hours = (3C1*3C1*3C1)/48 = 9/16

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11 Jul 2026, 00:33

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The cockpit crew for a long-haul flight must consist of 1 pilot and 2 copilots, selected from an available pool of 4 pilots and 6 copilots. One of the pilots, Hugo, is only certified to serve as pilot if each of the two selected copilots has at least 500 hours of flight experience, a qualification held by 3 of the 6 available copilots. The remaining pilots can fly with any copilot. A cockpit crew will be selected at random from all possible crew combinations that meet these conditions.

In order to find no. of valid crews.
1. No. of valid crew with Hugo: 1×3C1 = 3
2. No. of valid crews w/o Hugo: 3C1×6C2 = 45
So total no. Of possible valid crews = 45 + 3 = 48

In case Hugo is pilot,
No. of valid crews = 3
So probability of valid crew with pilot Hugo = 3/48 = 1/16

In case exactly one copilot has 500 hrs flight experience,
No. Of valid crews = 3C1×3C1×3C1 = 27
So, probability or valid crew with exactly one copilot having 500 hrs experience = 27/48 = 9/16

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11 Jul 2026, 00:58

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This question requires a clear understanding of Permutation and Combination concept along with Probability.

How I went about it is in this way:

Probability = Favourable / Total Outcomes

Total outcomes first (i) When Hugo is the Pilot; (ii) When Any other Pilot other than Hugo;
Hugo - 1C1 X 3C2 = 1 X 3 = 3
Any other - 3C1 X 6C2 = 3X 15 = 45
Total outcomes = 48

Now Probability that Hugo is the Pilot is 1C1 AND Co-Pilot >=500 Hrs AND Co-Pilot >=500 Hours
So that can be represented as 1C1 X 3C2 [3 Co-Pilots having >=500 Hrs and 2 must be selected].

Thus Hugo = 1C1 X 3C2 / 48 = 3/48 or 1/16

Similarly now for the Second Set:

We need to select 1 Pilot and 2 Co-Pilots where exactly 1 Co-Pilot has less than 500 hours. We know 3 have >=500 hours so that means 3 others have <500 hours.

So Hugo cannot be the Pilot because he needs both to have >=500 Hours, thus one of the other 3 must be the Pilot.

So that's 3C1 X 3C1 [Exactly 1 Co-Pilot <500 Hrs] X 3C1 [The other Co-Pilot >=500 Hours] / 48 = 27/48 or 9/16

Tricky question but requires a careful understanding of the concepts and then applying them in terms of what is required. More practice sets of this kind would be very helpful.

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11 Jul 2026, 01:25

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Crew must have 1 Pilot and 2 copilot.
Total pilots= 4 and total copilots= 6
Hugo can only serve as pilot if two copilots have atleast 500 hours of flight experience.
3 copilots have more than 500 hours of flight experience.

Total crews that can be formed with Hugo as pilot= 1*3C2 =1*3=3
Total crews that can be formed with other 3 memebers as pilots = 3*6C2 =3*15=45
Total crews possible with given constraints = 3+45=48

Probability that Hugo is the pilot= Crews when Hugo is pilot/ Total crews= 3/48=1/16

Probability that exactly 1 of the 2 copilots on crew has less than 500 hours,
Total required copilot pairs= 1 from 3 copilots who have less than 500 hours experience + 1 from 3 copilots who have more than 500 hours flight experience = 3C1*3C1=3*3=9
Since 1 copilot has less than 500 hours flight experience, Hugo cannot be chosen as pilot. Pilots can be chosen in 3 ways.
Required crew can be selected in 3*9= 27 ways.
Probability that exactly 1 of 2 chosen copilots have less than 500 hrs flight experience = 27/48=9/16

Hugo is the pilot= 1/16
Exactly 1 copilot has less than 500 hours=9/16

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11 Jul 2026, 01:40

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Bunuel

1st part ---> hugo is pilot. then 2 co pilots have to from 3 experienced pilots. 3C2 combinations = 3. for remaining 3 pilots, 6C2 = 15 for each. total = 45+3 =48. probability for hugo = 3/48 = 1/16.

2nd part ---> exactly one copilot has less than 500hours. with hugo its not possible. for other 3, (3c1x3c1) = 9 for each. total 27. thus, probability = 27/48 = 9/16.

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11 Jul 2026, 02:11

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Let's first count the total number of valid cockpit crews.

There are 4 pilots, including Hugo, and 6 copilots.

Among the 6 copilots:
3 have at least 500 hours of flight experience
3 have less than 500 hours

Case 1: Hugo is the pilot

Hugo can fly only if both selected copilots are experienced.

So we choose 2 copilots from the 3 experienced copilots:

C(3,2) = 3

So there are 3 valid crews with Hugo as the pilot.

Case 2: One of the other 3 pilots is selected

The other pilots have no restrictions and can fly with any pair of copilots.

Number of ways to choose 2 copilots from 6:

C(6,2) = 15

Since there are 3 such pilots:

3 × 15 = 45 valid crews.

Therefore, the total number of valid crews is:

3 + 45 = 48

Probability that Hugo is the pilot

There are 3 valid crews with Hugo as the pilot out of 48 total valid crews.

Probability = 3/48 = 1/16

Answer: 1/16

Probability that exactly one copilot has less than 500 hours

To have exactly one less-experienced copilot, choose:

- 1 experienced copilot from 3 = 3 ways
- 1 less-experienced copilot from 3 = 3 ways

Total copilot pairs: 3 × 3 = 9

Hugo cannot fly with these pairs because he requires both copilots to be experienced.

So only the other 3 pilots can be paired with these 9 copilot pairs.

Number of valid crews: 3 × 9 = 27

Therefore, probability = 27/48 = 9/16

Answer: 9/16

Hugo is the pilot = 1/16
Exactly 1 copilot has less than 500 hours = 9/16

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11 Jul 2026, 02:28

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Probability Hugo is pilot = 1/16
Probability that exactly 1 co pilot has experience less than 500 hrs = 9/16

sol
1 P + 2 C = N

Available pool - 4P + 6C

Hugo = Pilot when 2C >= 500 hrs

Only 3 have experience => 500 hrs
3 have experience < 500 hrs

So When Hug0 is pilot , the total possible outcomes = ( 3 = 3 outcomes of copilot
2)
Hugo can form crew = 1x 3 =3

When Hugo is not pilot , the total possible outcomes = (6 = 6!/2!x4! = 15 outcomes
2)

Crew without hugo = 3 x 15 =45

Total possible outcomes of crew = 45+3 = 48

1) When hugo is pilot = 3/48 = 1/16

2) When exactly 1 co pilot has experience less than 500 hrs =
1 with experience = 3 choices
1 without experience = 3 choices
Total = 9

Hugo cannot be pilot so total pilot choices = 9x 3 = 27
P = 27/48 = 9/16

Bunuel

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11 Jul 2026, 02:42

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There are 4 pilots and 6 Co Pilots.
Out of the copilots we have 3 experienced (E) and 3 beginners (B).
Since a crew can have 1 Pilot + 2 Copilot:
Lets find out total Number of Crews:

Out of the 3 experienced Copilots we can choose 3C2 ways = 3
And in that case Hugo can fly in 3 different ways.

If the other three pilots are selected they can fly with any two copilots.
So 3 x 6C2 = 3 x 15 = 45
Total number of crews = 48.

From here if hugo is the pilot probability is : 3/48 = 1/16.... For Column 1.

Now if we have to choose 1 experienced and 1 beginner: Number of such pairs = 3 x 3 = 9.
Hugo is not qualified to be in such a crew so for the other 3 pilots the crew can be: 3 x 9 = 27.
=27/48 = 9/16.
This should be the answer.

Bunuel

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11 Jul 2026, 03:05

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Formula
robability = favorable outcome / total possible outcomes

First, calculate the total possible outcomes
1. Case Hugo (special case)
1 Pilot x 3 Co-pilot (>500h) = 3

2. Case Other Pilot
3 Pilot x 6 C 2 Co-pilot = 3 x 15 = 45

Total possible outcome: 3 + 45 = 48

Second, calculated the favourable outcome for specific case

1. Case Hugo
Requirement: mandatory paired with 2 copilots with >500h
1 pilot x 3 C 2 Co-pilot = 1 x 3 = 3

So, 1st Probability 3/48 = 1/16

2. Case Other Pilot: Exactly 1 copilot with <500h

Requirement: paired with 2 copilots
1 for >500h from 3 options
1 for <500h from 3 options

So for Co Pilot: 3C1 X 3C1 = 9

Pilot:
3 Pilot other than Hugo
3 C 1 = 3

Favorable outcome: 9 x 3 = 27

So, 2nd probability 27/48 = 9/16

Bunuel

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11 Jul 2026, 04:26

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The answer is Hugo is the pilot 1/16
Exactly one copilot has less than 500 hours
9/16
A crew equals one pilot +2 copilots. We count all the cruise that are actually allowed.
If Hugo is the pilot, then both copilots must be from the three with more than or equal 500 hours. Choosing two of those three gives three crews.
If the pilot is one of the other three pilots: any two of the six copilots work. That’s 15 copilot pairs each. So 3x15=45 crews.
Total allowed crews=3+45=48. The crew is picked at a random from these 48. Hugo is the pilot.
Three of the 48 crews have Hugo, so the probability is 3/48=1/16.
Exactly one copilot has less than 500 hours.
Three copilots have at least 500 hours, three have less than 500.
Hugo’s crew can never include a copilot under 500 hours so they contribute zero.
For each of the other three pilots pick one copilot from the qualified group(3 ways) and one from the under-500 group(3ways), which gifts nine pairs. So 3x9=27 crews.
Probability=27/48=9/16.

Bunuel

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11 Jul 2026, 04:58

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Total cases= special hugo case+non hugo case
hugo case- hugo is pilot, both copilot to be experienced
1c1x3c2=3
non hugo case- non hugo pilot, any copilot
3c1x6c2=45

Total cases=3+45=48

P(hugo pilots)=3/48=1/16

exactly one copilot <500 hrs- 3c1x3c1x3c1=27
P(one copilot<500hrs)=27/48=9/16

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11 Jul 2026, 05:20

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Total no of. pilot=4
total no of co pilots=6
no of pilots compatible with Hugo=3

For left col.
No of ways Hugo can be the pilot is= 3C2=3 ways
Total number of valid cockpit crews,
Case1. hugo is pilot=3 ways
case 2 Hugo is not pilot= 3C1*6C2=3*6*5/2=45
.: Total number of valid cockpit crews=3+45=48
.: probality of Hugo being the pilot is=3/48=1/16

for right col.,
The no of the case is=3C1* 3C1*3C1=27
.: In this case the probability=27/48=9/16

Bunuel

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11 Jul 2026, 05:45

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	Pilot	Co pilot
Required crew	1	2
Available crew	4	6
	Hugo + 3	3>500 hours
		3<500 hours

Total Outcome(TO) - C1 + C2

C1 - Hugo is flying
two co pilots can be chosen in 3C2 ways = 3

C2 - Hugo won't be flying
1 pilot can be chosen in 3C1 ways = 3
2 Co pilots can be chosen in 6C2 ways = 15

Total C2 = 3 XOXO 15 = 45

TO = 45 + 3 = 48

P(H) = 3/ 48 = 1/16 answer

P(exactly 1 co pilot with <500 hours) =\(\frac{ (1 pilot) * (1 Copilot <500 hours )* (1 Copilot >500 hours)}{45} \)= \(\frac{3C1 * 3C1 *3C1}{45} \)= \(\frac{27 }{ 45}\) = 9/16

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11 Jul 2026, 05:50

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Total possible combinations

group of 3 1 pilot and 2 co piolts

1 hugo as piolt => 1 * cobination of 2 out of 3 co piolts with 500hr+ exp => 3 possible combination
3 other piolts => 3 * combination of 2 from a group of 6 copiolts => 6C2 => 14 => 3*15 = 45 combinations

Total possible combinations = 48

first part: 3/48 = 1/16 probability
second part: one piolt with a experienced co and one non experienced co => 3 * (1 out of experienced 3 * 1 out of in experinecd 3) => 3*3*3 = 27
=> 27/48 => 9/16

Bunuel

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11 Jul 2026, 05:59

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The correct answer is
Hugo is the pilot: 1/16
Exactly 1 co pilot has less than 500 hrs= 9/16

here, from the question
4 pilots (Hugo + 3 other)
6 co-pilots (3 with >= 500 hrs and other 3 with less than 500 hrs)

Case 1, Hugo is the pilot. There is only 1 way possible for this.
Now for co pilots in this situation
3C2= 3 ways.
hence total crew= 3

Case 2, one other pilot is chosen (3 ways)
Co-pilots- 6C2 = 15
Total crew with Hugo = 3*15= 45
So, total valid crew= 3+45= 48.

Calculating the probability where Hugo is the pilot
no. of valid crews= 3
total valid crew = 48
Probability= 3/48=1/16

Calculating the probability where exactly 1 co-pilot has less than 500 hrs.
here, we need to check the valid crews that include exactly 1 co pilot with less than 500 hours.
From Hugo's crew: 0 crews fit the description, as all of Hugo's co-pilots are more than or equal to 500 hours.

From the other pilot's crew (45)
We must choose 1 co pilot with <500 hours and the 1 co pilot with >= 500 hours.

Co-pilot combinations- 3C1*3C1 = 9 ways
Pilot combination- 3
total= 3*9= 27 combinations

Probability= 27/48=9/16

Hence our answer.

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11 Jul 2026, 06:27

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Total:
+ Hugo: 3C2 = 3
+ The other pilots: 3*6C2 = 3*15 = 45
Total = 3+45 = 48

Hugo is the pilot:
3/48 = 1/16

Exactly 1 copilot has less than 500 hours:
If 3 copilots has at least 500 hours of flight experience, 6-3=3 copilots has less than 500 hours.
We can select each of the 3 copilots with each copilot more experienced: 3*3 = 9
Hugo cannot be the pilot: 3 pilots available.
3*9 = 27
27/48 = 9/16

Hugo is the pilot = 1/16
Exactly 1 copilot has less than 500 hours = 9/16