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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.
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.
_________________
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.
| Hugo is the pilot | Exactly 1 copilot has less than 500 hours | |
| \(\frac{1}{16}\) | ||
| \(\frac{3}{16}\) | ||
| \(\frac{1}{4}\) | ||
| \(\frac{9}{16}\) | ||
| \(\frac{19}{24}\) | ||
| \(\frac{15}{16}\) |
|
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Manhattan Prep Official Explanation
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.
Step 1: Understand the Prompt and Question
Glance at the answers. The fractions point to a Quant-based question. Further, the words probability and combinations in the question stem indicate that this is a Probability and Combinatorics problem. Finally, read the question stem. The task requires evaluating independent column conditions rather than a single joint event, meaning you can solve for the two columns separately.
Next, read the entire prompt and jot down the given information on your scratch paper:
Total available: 4 pilots, 6 copilots
Crew required: 1 pilot, 2 copilots
Certification: Hugo (pilot) can only fly with experienced copilots (at least 500 hours)
Copilots: 3 have at least 500 hours, 3 have less than 500 hours
Step 2: Plan your Approach
The specific details suggest listing scenarios when constraints are imposed (e.g. when Hugo is the pilot, as this limits the choice of copilots and makes the list short) and using formulaic combinatorics when there are minimal constraints other than choosing k from n (e.g. when Hugo is not the pilot). Calculate the number of possible crews for these two mutually exclusive scenarios (Hugo is pilot or not); their sum is the total number of possible crew combinations.
The standard combination formula for n choose k \(= \frac{n!}{k!(n-k)!}\) applies when order doesn’t matter, as it doesn’t for the two copilots in this problem. Where possible, check any formula result with logic.
Step 3: Solve the Problem
First determine the denominator of both answers, which is the total number of possible crew combinations that meet the given constraints.
If Hugo is the pilot: Choose Hugo and then choose 2 of the 3 copilots who have at least 500 hours flight experience. If the 3 copilots with enough experience are A, B, and C, the possible crews are:
Hugo, A, B
Hugo, A, C
Hugo, B, C
Alternatively, apply the formula:
Crews with Hugo = 1 * 3!/(2!1!) = 3
If Hugo is not the pilot, choose 1 of the other 3 pilots, and any 2 of the 6 available copilots:
Crews without Hugo = 3!/(1!2!) * 6!/(2!4!) = 45
Logical check: There are 3 unique ways to choose the non-Hugo pilot, so the 3 term makes sense. Of the 6 copilots, the first copilot can pair with any of the other 5. Then for the second copilot, the pairing with the first has already been counted, so count only the pairing of the second copilot with any of the other 4, and so on. There are 5 + 4 + 3 + 2 + 1 + 0 = 15 unique pairs of copilots that can be made from the 6 copilots available.
In total, there are 45 + 3 = 48 possible crew combinations. This is the denominator for both of the probabilities in the question.
Solve for Column 1, in which Hugo is the pilot. As shown above, Hugo is the pilot in exactly 3 of the 48 possible crew combinations:
Probability = 3/48 = 1/16
Solve for Column 2: Exactly 1 copilot has less than 500 hours implies that the crew must have exactly 1 copilot from the experienced group (3 choices) and exactly 1 copilot from the less-experienced group (3 choices).
# of such copilot pairs = 3!/(1!2!) * 3!/(1!2!) = 9
Hugo cannot fly with any of these pairs because he requires both copilots to have at least 500 hours of flight experience, but the other 3 pilots can fly with any of these pairs.
# of resulting permissible crews = 3 pilots ✕ 9 copilot pairs = 27
Probability = 27/48 = 9/16
GMAT-Club-Forum-zfovdxvn.png [ 17.48 KiB | Viewed 1545 times ]
_________________
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.
Step 1: Understand the Prompt and Question
Glance at the answers. The fractions point to a Quant-based question. Further, the words probability and combinations in the question stem indicate that this is a Probability and Combinatorics problem. Finally, read the question stem. The task requires evaluating independent column conditions rather than a single joint event, meaning you can solve for the two columns separately.
Next, read the entire prompt and jot down the given information on your scratch paper:
Total available: 4 pilots, 6 copilots
Crew required: 1 pilot, 2 copilots
Certification: Hugo (pilot) can only fly with experienced copilots (at least 500 hours)
Copilots: 3 have at least 500 hours, 3 have less than 500 hours
Step 2: Plan your Approach
The specific details suggest listing scenarios when constraints are imposed (e.g. when Hugo is the pilot, as this limits the choice of copilots and makes the list short) and using formulaic combinatorics when there are minimal constraints other than choosing k from n (e.g. when Hugo is not the pilot). Calculate the number of possible crews for these two mutually exclusive scenarios (Hugo is pilot or not); their sum is the total number of possible crew combinations.
The standard combination formula for n choose k \(= \frac{n!}{k!(n-k)!}\) applies when order doesn’t matter, as it doesn’t for the two copilots in this problem. Where possible, check any formula result with logic.
Step 3: Solve the Problem
First determine the denominator of both answers, which is the total number of possible crew combinations that meet the given constraints.
If Hugo is the pilot: Choose Hugo and then choose 2 of the 3 copilots who have at least 500 hours flight experience. If the 3 copilots with enough experience are A, B, and C, the possible crews are:
Hugo, A, B
Hugo, A, C
Hugo, B, C
Alternatively, apply the formula:
Crews with Hugo = 1 * 3!/(2!1!) = 3
If Hugo is not the pilot, choose 1 of the other 3 pilots, and any 2 of the 6 available copilots:
Crews without Hugo = 3!/(1!2!) * 6!/(2!4!) = 45
Logical check: There are 3 unique ways to choose the non-Hugo pilot, so the 3 term makes sense. Of the 6 copilots, the first copilot can pair with any of the other 5. Then for the second copilot, the pairing with the first has already been counted, so count only the pairing of the second copilot with any of the other 4, and so on. There are 5 + 4 + 3 + 2 + 1 + 0 = 15 unique pairs of copilots that can be made from the 6 copilots available.
In total, there are 45 + 3 = 48 possible crew combinations. This is the denominator for both of the probabilities in the question.
Solve for Column 1, in which Hugo is the pilot. As shown above, Hugo is the pilot in exactly 3 of the 48 possible crew combinations:
Probability = 3/48 = 1/16
Solve for Column 2: Exactly 1 copilot has less than 500 hours implies that the crew must have exactly 1 copilot from the experienced group (3 choices) and exactly 1 copilot from the less-experienced group (3 choices).
# of such copilot pairs = 3!/(1!2!) * 3!/(1!2!) = 9
Hugo cannot fly with any of these pairs because he requires both copilots to have at least 500 hours of flight experience, but the other 3 pilots can fly with any of these pairs.
# of resulting permissible crews = 3 pilots ✕ 9 copilot pairs = 27
Probability = 27/48 = 9/16
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General Discussion
in general the total ways would be - 4c1 * 6c2 = 4 * 15 = 60
but we have a restriction with Hugo, so for Hugo only 3C2 Copilots are valid = 3
and for the rest 3 pilots there is no restriction - 3c1 * 6c2 = 3 * 15 = 45
so Total ways = 48
I calculated first without the hugo restriction and ended up with random answer (3/60), then the above occurred to me ...haha
So for First Question - P(H) = 3/48 = 1/16
and for second question P(only one >500) =
for Copilot - 3C1 * 3C1 = 9
for pilot - 3C2 (as Hugo can't be accepted)
so, P (Only one >500) = 9*3 / 48 = 9/16
but we have a restriction with Hugo, so for Hugo only 3C2 Copilots are valid = 3
and for the rest 3 pilots there is no restriction - 3c1 * 6c2 = 3 * 15 = 45
so Total ways = 48
I calculated first without the hugo restriction and ended up with random answer (3/60), then the above occurred to me ...haha
So for First Question - P(H) = 3/48 = 1/16
and for second question P(only one >500) =
for Copilot - 3C1 * 3C1 = 9
for pilot - 3C2 (as Hugo can't be accepted)
so, P (Only one >500) = 9*3 / 48 = 9/16
Posts: 60
i am going with 1/16 as hugo is the pilot and with 9/16 as exactly 1 copilot has less than 500 hours.
total 4 pilots, copilots 6 in total
case 1: hugo is the pilot- the 2 copilots must both be exp, choose 2 out of 3 = 3 ways
case 2: hugo is not the pilot - 2 copilots out of 6 available = 15 ways
total scenario = 3x15 = 45, total valid crew = 3+45 = 48 ways
prob hugo is pilot = 3/48 = 1/16
prob exactly 1 copilot has less than 500 hours - 27/48 = 9/16
total 4 pilots, copilots 6 in total
case 1: hugo is the pilot- the 2 copilots must both be exp, choose 2 out of 3 = 3 ways
case 2: hugo is not the pilot - 2 copilots out of 6 available = 15 ways
total scenario = 3x15 = 45, total valid crew = 3+45 = 48 ways
prob hugo is pilot = 3/48 = 1/16
prob exactly 1 copilot has less than 500 hours - 27/48 = 9/16
Ways for ONLY Hugo = 1 x 3C2 (copilots with 500+ hours) = 1 x 3 = 3
Ways for others = 3C1 x 6C2 = 3 x 15 = 45
Total = 45 + 3 = 48
P(Hugo) = 3/48 = 1/16
Ways for ONLY 1 copilot with 500+ = 3C1 (copilots 500+) x 3C1 (copilots less than 500) x 3C1 (Pilots) = 27
P(ONLY 1 Copilot with 500+) = 27/48 = 9/16
Ways for others = 3C1 x 6C2 = 3 x 15 = 45
Total = 45 + 3 = 48
P(Hugo) = 3/48 = 1/16
Ways for ONLY 1 copilot with 500+ = 3C1 (copilots 500+) x 3C1 (copilots less than 500) x 3C1 (Pilots) = 27
P(ONLY 1 Copilot with 500+) = 27/48 = 9/16
Answer: Hugo is the pilot = 1/16, Exactly 1 co-pilot less than 500 hrs = 9/16
Hugo is the pilot:
Total number of desirable outcomes = all outcomes where there are two co-pilots with >500 hrs = 1C1 * 3C2 = 3
Total number of possible outcomes = 4C1 * 6C2 - Invalid crews (hugo is pilot but co-pilots don't both have >500 hrs) = 60 - 12 = 48
3/48 = 1/16
Exactly 1 co-pilot less than 500 hrs:
Total number of desirable outcomes = 3 possible pilots * 3 possible experienced co-pilots * 3 possible less experienced co-pilots = 27
Total number of possible outcomes = 48
27/48 = 9/16
Hugo is the pilot:
Total number of desirable outcomes = all outcomes where there are two co-pilots with >500 hrs = 1C1 * 3C2 = 3
Total number of possible outcomes = 4C1 * 6C2 - Invalid crews (hugo is pilot but co-pilots don't both have >500 hrs) = 60 - 12 = 48
3/48 = 1/16
Exactly 1 co-pilot less than 500 hrs:
Total number of desirable outcomes = 3 possible pilots * 3 possible experienced co-pilots * 3 possible less experienced co-pilots = 27
Total number of possible outcomes = 48
27/48 = 9/16
total possible crews =
When Hugo is pilot = 1c1 * 3c2 = 3
When Hugo is not pilot = 3c1* 6c2 = 45
Total = 48
Now P that hugo is pilot = 1c1*3c2 / 48 = 3/48 = 1/16
When 1 copilot is experienced one isnt = (hugo cant be pilot now) = 3c1*3c1*3c1 / 48 = 27/48 = 9/16
When Hugo is pilot = 1c1 * 3c2 = 3
When Hugo is not pilot = 3c1* 6c2 = 45
Total = 48
Now P that hugo is pilot = 1c1*3c2 / 48 = 3/48 = 1/16
When 1 copilot is experienced one isnt = (hugo cant be pilot now) = 3c1*3c1*3c1 / 48 = 27/48 = 9/16
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Posts: 317
Let copilot be
3 e [>500]
3 l [<500]
Total Valid crews:
Hugo as pilot: choose 3 e = c(3,2) = 3
other 3 pilots: can fly with any 2 of 6 copilots = 3 * c(6,2)
= 3*15 = 45
total valid crews = 3+45 = 48
1. Prob Hugo is pilot:
= 3/48
= 1/16
2. Prob exactly 1 copilot l[<500]
choose 1 exp and 1 less exp copilot
c(3,1) * c(3,1) = 3*3 = 9
only the other 3 pilots can fly with this combinations
3*9 = 27
Prob = 27/48 = 9/16
3 e [>500]
3 l [<500]
Total Valid crews:
Hugo as pilot: choose 3 e = c(3,2) = 3
other 3 pilots: can fly with any 2 of 6 copilots = 3 * c(6,2)
= 3*15 = 45
total valid crews = 3+45 = 48
1. Prob Hugo is pilot:
= 3/48
= 1/16
2. Prob exactly 1 copilot l[<500]
choose 1 exp and 1 less exp copilot
c(3,1) * c(3,1) = 3*3 = 9
only the other 3 pilots can fly with this combinations
3*9 = 27
Prob = 27/48 = 9/16
Posts: 49
P(Hugo)=1/4
p(Copilot>=500)=3/6 =1/2 p(copilot<500)=1/2
Hugo cant fly without both copilots>=500 hrs of flying experience
P(1 copilot <500 and 1 copilot>=500)=3/4*1/2*1/2=3/16
p(Copilot>=500)=3/6 =1/2 p(copilot<500)=1/2
Hugo cant fly without both copilots>=500 hrs of flying experience
P(1 copilot <500 and 1 copilot>=500)=3/4*1/2*1/2=3/16
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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.
Hugo can be pilot if 2 co pilot have > 500 hours flight experience , which 3 out 6 have
3c2 , 3 ways to select that co pilot
total combination to select co pilot
6c2 = 15 ways
there are 3 other pilots we have total of
15*3 45 ways
total combination will be 45+3 ; 48
hugo is pilot P will be = 3 /48 ;
1/16 is correct
for second case s probability that exactly 1 of the 2 copilots < 500 hours
total ways will be
total 3 pilots * 3c1 ( pilots with experience ) * 3c1 ( pilots who have no experience)
3c1 * 3c1 * 3c1 = 27 ways
27/48 ; 9/16
1/16 & 9/16 are correct options
Hugo can be pilot if 2 co pilot have > 500 hours flight experience , which 3 out 6 have
3c2 , 3 ways to select that co pilot
total combination to select co pilot
6c2 = 15 ways
there are 3 other pilots we have total of
15*3 45 ways
total combination will be 45+3 ; 48
hugo is pilot P will be = 3 /48 ;
1/16 is correct
for second case s probability that exactly 1 of the 2 copilots < 500 hours
total ways will be
total 3 pilots * 3c1 ( pilots with experience ) * 3c1 ( pilots who have no experience)
3c1 * 3c1 * 3c1 = 27 ways
27/48 ; 9/16
1/16 & 9/16 are correct options
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Cockpit crew : 1 pilot + 2 copilots
Availability : 4 pilots & 6 copilots
Hugo (pilot) - > Each of 2 copilots has > 500 hours of flight experience ( 3 out of 6 copilots have this qualification)
Remaining 3 pilots - > Any 2 copilots out of 6.
Total selections possible = 1*3C2 + 3C1*6C2 = 1*3 + 3*15 = 48
The number of selections with Hugo as pilot = 1*3C2 = 3
The probability that Hugo is a pilot on the selected crew = 3/48 = 1/16
The number of selections with exactly 1 copilot has < 500 hours experience = 3C1 * 3C1 * 3C1 = 27
The probability that exactly 1 copilot has < 500 hours experience = 27/48 = 9/16
Availability : 4 pilots & 6 copilots
Hugo (pilot) - > Each of 2 copilots has > 500 hours of flight experience ( 3 out of 6 copilots have this qualification)
Remaining 3 pilots - > Any 2 copilots out of 6.
Total selections possible = 1*3C2 + 3C1*6C2 = 1*3 + 3*15 = 48
The number of selections with Hugo as pilot = 1*3C2 = 3
The probability that Hugo is a pilot on the selected crew = 3/48 = 1/16
The number of selections with exactly 1 copilot has < 500 hours experience = 3C1 * 3C1 * 3C1 = 27
The probability that exactly 1 copilot has < 500 hours experience = 27/48 = 9/16
| Hugo is the pilot | Exactly 1 pilot has less than 500 hours |
| 1/16 | 9/16 |
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.
Total value crews with Hugo as Pilot= 1*3C2 = 3
Total valid crews with other 3 pilots = 3*6C2= 45
Total valid cases= 3+45= 48
1. When Hugo is the Pilot = 3/48= 1/16
2. Exactly 1 copilot has less than 500 hours.
Needed Copilot pairs= 3C1*3C1 =9
Except Hugo other 3 pilots can be chosen = 3
Crew can be selected=9*3= 27
Required probability= 27/48= 9/16
Total value crews with Hugo as Pilot= 1*3C2 = 3
Total valid crews with other 3 pilots = 3*6C2= 45
Total valid cases= 3+45= 48
1. When Hugo is the Pilot = 3/48= 1/16
2. Exactly 1 copilot has less than 500 hours.
Needed Copilot pairs= 3C1*3C1 =9
Except Hugo other 3 pilots can be chosen = 3
Crew can be selected=9*3= 27
Required probability= 27/48= 9/16
Posts: 81
Our first step in this question would be to make total number of ways of choosing different crew.
For total ways of different crew:
Case 1; when Hugo is the pilot, we would have only 3 copilots to choose(as given)
1 way to choose Hugo and 3C2 ways to choose copilot
1 * 3 = 3 ways
Case 2: when Hugo is not the pilot:
3C1 ways to choose pilot and 6C2 (=15) ways to choose copilot
3 * 15 = 45 ways
Total ways to choose different crews = 3 + 45 = 48 ways
(I) Now P(Hugo is the pilot) = 3/48 = 1/16
(II) P(Exactly 1 copilot has less than 500 hours) - Hugo's case would be eliminated as he requires both copilots to have at least 500 hours of flight experience.
Ways to make different crews with 1 copilot with less than 500 hours of flight experience:
Choose pilot(3C1), Choose one copilot of the 3 with at least 500 hours of flight experience(3C1) and choose one copilot with less than 500 hours of flight experience(3C1)
= 3 * 3 * 3 = 27 = total ways to make different crews with less than 500 hours of flight experience
P(Exactly 1 copilot has less than 500 hours of flight experience) = 27/48 = 9/16
(I) 1/16
(II) 9/16
For total ways of different crew:
Case 1; when Hugo is the pilot, we would have only 3 copilots to choose(as given)
1 way to choose Hugo and 3C2 ways to choose copilot
1 * 3 = 3 ways
Case 2: when Hugo is not the pilot:
3C1 ways to choose pilot and 6C2 (=15) ways to choose copilot
3 * 15 = 45 ways
Total ways to choose different crews = 3 + 45 = 48 ways
(I) Now P(Hugo is the pilot) = 3/48 = 1/16
(II) P(Exactly 1 copilot has less than 500 hours) - Hugo's case would be eliminated as he requires both copilots to have at least 500 hours of flight experience.
Ways to make different crews with 1 copilot with less than 500 hours of flight experience:
Choose pilot(3C1), Choose one copilot of the 3 with at least 500 hours of flight experience(3C1) and choose one copilot with less than 500 hours of flight experience(3C1)
= 3 * 3 * 3 = 27 = total ways to make different crews with less than 500 hours of flight experience
P(Exactly 1 copilot has less than 500 hours of flight experience) = 27/48 = 9/16
(I) 1/16
(II) 9/16
Posts: 112
4 pilots (including Hugo)
6 copilots ( 3 with over 500 hours, 3 less than 500)
Hugo can only fly with two experienced copilots
Number of crews is 3 choose 2 equals 3
Other pilots can have any copilots pair. 6 choose 2 equals 15
therefore three pilots can have 3 x 15 = 45 possibilities
Therefor total number of valid crews is
3 + 45 = 48
If Hugo is pilot than probability is
3/48 or 1/16
Exactly one copilot has less than 500 hours
1 less than 500 hours from 3
1 more than 500 hours from 3
number of pairs is 3x3=9
these pairs can not fly with Hugo
number of valid crews with exactly one pilots with under 500 hours is
9 x 3 = 27
27/48 = 9/16
6 copilots ( 3 with over 500 hours, 3 less than 500)
Hugo can only fly with two experienced copilots
Number of crews is 3 choose 2 equals 3
Other pilots can have any copilots pair. 6 choose 2 equals 15
therefore three pilots can have 3 x 15 = 45 possibilities
Therefor total number of valid crews is
3 + 45 = 48
If Hugo is pilot than probability is
3/48 or 1/16
Exactly one copilot has less than 500 hours
1 less than 500 hours from 3
1 more than 500 hours from 3
number of pairs is 3x3=9
these pairs can not fly with Hugo
number of valid crews with exactly one pilots with under 500 hours is
9 x 3 = 27
27/48 = 9/16
Posts: 16
Hugo as a Pilot - condition - both Copilot > 500 exp
Not Hugo as a Pilot - condition - anyone can be a copilot.
So, case 1 - Hugo as a Pilot - the other 2 can be 3c2 = 1(Hugo)*3 (out of 3 exp we can select 2)
case 2 - Not Hugo as a Pilot - space can be filled as - 3c1(pilots) * 6c2(other two copilots) = 45
Total. combination - 45 + 3 = 48
1. Now, Hugo Pilot - prob - 3/48 = 1/16
2. exactly one copilot with less than 500 exp
In this hugo can't be the pilot(since he needs both copilot > 500 exp)
So, the remaining pilot can be 3, and for the copilot, we need to pick exactly 1 with less than 500 exp - so ->
combination - 3(no of remaining pilot) * 3c1(one with > 500 exp) * 3c1(one with < 500 exp) = 27
prob - 27/48 = 9/16
So, answer = 1/16, 9/16
Not Hugo as a Pilot - condition - anyone can be a copilot.
So, case 1 - Hugo as a Pilot - the other 2 can be 3c2 = 1(Hugo)*3 (out of 3 exp we can select 2)
case 2 - Not Hugo as a Pilot - space can be filled as - 3c1(pilots) * 6c2(other two copilots) = 45
Total. combination - 45 + 3 = 48
1. Now, Hugo Pilot - prob - 3/48 = 1/16
2. exactly one copilot with less than 500 exp
In this hugo can't be the pilot(since he needs both copilot > 500 exp)
So, the remaining pilot can be 3, and for the copilot, we need to pick exactly 1 with less than 500 exp - so ->
combination - 3(no of remaining pilot) * 3c1(one with > 500 exp) * 3c1(one with < 500 exp) = 27
prob - 27/48 = 9/16
So, answer = 1/16, 9/16
Posts: 28
I have taken a slightly rudimentary approach as I have not finished studying this topic.
Happy to see the best method. here is how I did it:
We have four pilots, Hugo + 3 others.
Let's call them A, B, C, H (Hugo)
We have 6 co-pilots, let's call them:
1,2,3,4,5,6
Out of these, only three have 500+ hours of flying experience (Let them be 1,2,3)
Since we need to find the probabilbity of Hugo being the pilot, let's look at all possible combinations.
First for 'A':
A (paired with) 12 | 13 | 14 | 15 | 16 |
A (paired with) 23 | 24 | 25 | 26 |
A (paired with) 34 | 35 | 36 |
A (paired with) 45 | 46
A (paired with) 56
Total possibilites for A- 15.
Similarly, since B & C can also fly with any copilot, pairings for them also will be 15 each.
For H-
H (paired with) 12 | 13 |
H (paired with) 23 |
Total possibilites for H- 3
Total possibilities across pilots- 15+15+15+3= 48
Therefore probability of Hugo being the pilot- (3/48) = (1/16)
---
Now, let's look at the probability of Exactly one copilot with LESS than 500 hours.
Or in other words, with only one copilot with MORE THAN 500 hours.
If you see the table above, for A its possible only in 9 cases (14 | 15 | 16 | 24 | 25 | 26 | 34 | 35 | 36)
(Since we had assumed that 1,2 & 3 have over 500 hours of flying experience.
Similary for B & C as well there will be 9 cases each.
For Hugo, there will be no cases as BOTH copilots need to have over 500 hours of flying experience.
Hence, we get: (9+9+9)/48
=27/48
=9/16
Therefore the answers are:
| 1/16 | 9/16 |
Happy to see the best method. here is how I did it:
We have four pilots, Hugo + 3 others.
Let's call them A, B, C, H (Hugo)
We have 6 co-pilots, let's call them:
1,2,3,4,5,6
Out of these, only three have 500+ hours of flying experience (Let them be 1,2,3)
Since we need to find the probabilbity of Hugo being the pilot, let's look at all possible combinations.
First for 'A':
A (paired with) 12 | 13 | 14 | 15 | 16 |
A (paired with) 23 | 24 | 25 | 26 |
A (paired with) 34 | 35 | 36 |
A (paired with) 45 | 46
A (paired with) 56
Total possibilites for A- 15.
Similarly, since B & C can also fly with any copilot, pairings for them also will be 15 each.
For H-
H (paired with) 12 | 13 |
H (paired with) 23 |
Total possibilites for H- 3
Total possibilities across pilots- 15+15+15+3= 48
Therefore probability of Hugo being the pilot- (3/48) = (1/16)
---
Now, let's look at the probability of Exactly one copilot with LESS than 500 hours.
Or in other words, with only one copilot with MORE THAN 500 hours.
If you see the table above, for A its possible only in 9 cases (14 | 15 | 16 | 24 | 25 | 26 | 34 | 35 | 36)
(Since we had assumed that 1,2 & 3 have over 500 hours of flying experience.
Similary for B & C as well there will be 9 cases each.
For Hugo, there will be no cases as BOTH copilots need to have over 500 hours of flying experience.
Hence, we get: (9+9+9)/48
=27/48
=9/16
Therefore the answers are:
| 1/16 | 9/16 |
total valid combination = when the other 3 pilots are selected + when Hugo is selected = 3c1*6c2 + 3c2 = 45 + 3 = 48
Case 1: Hugo is the pilot:
2 out of 3 valide copilots can be onboarded. so 3c2 = 3
total = 48
so, P(1) = 3/48 = 1/16
Case 2: exactly 1 copilot with less that required exp.
We can ignore Hugo case as both should be more expd.
and we have 3 with less exp and 3 with more experience, so that combination will be
3c1(selecting one pilot out of 3) * 3c1(selecting 1 copilot with less exp)*3c1(selecting 1 copilot with more exp) = 27
so P(2) = 27/48 = 9/16
Case 1: Hugo is the pilot:
2 out of 3 valide copilots can be onboarded. so 3c2 = 3
total = 48
so, P(1) = 3/48 = 1/16
Case 2: exactly 1 copilot with less that required exp.
We can ignore Hugo case as both should be more expd.
and we have 3 with less exp and 3 with more experience, so that combination will be
3c1(selecting one pilot out of 3) * 3c1(selecting 1 copilot with less exp)*3c1(selecting 1 copilot with more exp) = 27
so P(2) = 27/48 = 9/16
Posts: 73
IMO : 1/16 for the 1st one and 9/16 for the 2nd one
although not so sure
although not so sure
| Pilot | copilot | combination possible | |
| typ1 t<500hr | 1 | 3 | 3 |
| typ2 | 3 | 6 | 3*6C2=3*15=45 |
| total | 48 |
Hugo is type 1 = P(H) =3/48=1/16
Prob of 1 of 2 copilots having t<500hrs means Hugo can be selected so combo possible is =3C1*3c1*3C1=27
P=27/48=9/16
Hugo is the only certified to serve as pilot if each of the two selected co pilots has at least 500 hours of flight experience.
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