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10 Jul 2026, 10:18
Kudos
Bookmarks
A tricky question..
So Hugo requires 2 out of 3 > 500 hours experienced copilot.
Total possible combo of 6 copilots, wherein given that each crew will consists of 2 copilot will be 6C2= 6!/(4!*2!)=15
Maximum possible combination of crew members if there will be no constraint of flying hours will be 4C1* 15=60
But Hugo can make combo with 3 copilot as they have >500 hours of flying experience. So his possible combinations are 3C2=3
and other 3 pilots can make crew combination with any 6 copilot. Total combo= 3C1*6C2=45
Total possible combinations actually are 45+3=48 and with Hugo only 3. so probability of Hugo combo in random selection=3/48=1/16
Second part:- P exactly one has experience less than500 hours in crew as copilot
This is possible with pilots except Hugo. so 3 pilots. Also out of 3 experience greater than 500 and 3 less than 500 we have to make crew combo of 2 copilots .
so, it is done with 3C1*3C1=9
So total combo possible = 9*3C1=27
Probability of this in random selection is 27/48=9/16
So Hugo requires 2 out of 3 > 500 hours experienced copilot.
Total possible combo of 6 copilots, wherein given that each crew will consists of 2 copilot will be 6C2= 6!/(4!*2!)=15
Maximum possible combination of crew members if there will be no constraint of flying hours will be 4C1* 15=60
But Hugo can make combo with 3 copilot as they have >500 hours of flying experience. So his possible combinations are 3C2=3
and other 3 pilots can make crew combination with any 6 copilot. Total combo= 3C1*6C2=45
Total possible combinations actually are 45+3=48 and with Hugo only 3. so probability of Hugo combo in random selection=3/48=1/16
Second part:- P exactly one has experience less than500 hours in crew as copilot
This is possible with pilots except Hugo. so 3 pilots. Also out of 3 experience greater than 500 and 3 less than 500 we have to make crew combo of 2 copilots .
so, it is done with 3C1*3C1=9
So total combo possible = 9*3C1=27
Probability of this in random selection is 27/48=9/16
Posts: 33
10 Jul 2026, 10:18
Kudos
Bookmarks
All valid combinations=(not Hugo)*(Any two copilots)+(Hugo)*(Two out of three copilots with experience)=3C1*6C2+1C1*3C2=45+3=48
Combinations where Hugo is the pilot=(Hugo)*(Two out of three copilots with experience)=1C1*3C2=3
Combinations where exactly one copilot has experience=(not Hugo)*(one copilot from three experienced)*(one copilot from three inexperienced)=3C1*3C1*3C1=27
P(Hugo is the pilot)=Combinations where Hugo is the pilot/All valid combinations=3/48=1/16
P(exactly one of two copilots has experience)=Combinations where exactly one copilot has experience/All valid combinations=27/48=9/16
Combinations where Hugo is the pilot=(Hugo)*(Two out of three copilots with experience)=1C1*3C2=3
Combinations where exactly one copilot has experience=(not Hugo)*(one copilot from three experienced)*(one copilot from three inexperienced)=3C1*3C1*3C1=27
P(Hugo is the pilot)=Combinations where Hugo is the pilot/All valid combinations=3/48=1/16
P(exactly one of two copilots has experience)=Combinations where exactly one copilot has experience/All valid combinations=27/48=9/16
10 Jul 2026, 10:29
Kudos
Bookmarks
We are given that,
Cockpit Crew is made up of 1 pilot and 2 co-pilots
There are 4 pilots and 6 co-pilots available
Hugo can only fly if both the co pilots with him have > 500 hrs of experience. There are 3 such pilots from the 6 pilots available.
Column 1: Probability of Hugo being the pilot
Hugo can only be the pilot if we choose both such co pilots who have an experience > 500 hours
Thus, we have to choose 2 out 3 available pilots 3C2 = 3
The only other possibility is that Hugo is not the pilot. That happens when one of the 3 other pilots are chose. Any 2 copiilots can be chosen as the other pilots have no constraint on who can be their co-pilots.
Thus, ways of choosing 1 pilot from 3 available pilots = 3C1 = 3
ways of choosing 2 co-pilots from 6 available co-pilots = 6C2 = 15
Thus, total combinations where Hugo is not the pilot = 3*15 = 45
Overall total number of combinations = combinations where Hugo is pilot (3) + Comibinations where he is not the pilot (45) = 48
Thus P (Hugo is pilot) = 3/48 = 1/16
Column 2: Exactly one of the two co pilots has experience > 500 hrs.
Since we only have to choose 2 co-pilots, one will have experience > 500 hrs and the other will not. We have 3 available in each category
Thus,
Ways of choosing a co pilot who has experience greater > 500 hrs = 3C1 = 3 and the same for the co pilot experience < 500 hrs = 3C1 = 3
Thus, total combinations of co pilots = 3*3 = 9
We will also have 3 pilots who can fly with these combinations of co pilots (Hugo cannot, since both are not experience > 500).
Thus, total number of combinations with 1 pilot and 2 co pilots, such that exactly one co pilot has experience > 500 hrs = 3*9 = 27
Hence, Probability = 27/48 = 9/16
Cockpit Crew is made up of 1 pilot and 2 co-pilots
There are 4 pilots and 6 co-pilots available
Hugo can only fly if both the co pilots with him have > 500 hrs of experience. There are 3 such pilots from the 6 pilots available.
Column 1: Probability of Hugo being the pilot
Hugo can only be the pilot if we choose both such co pilots who have an experience > 500 hours
Thus, we have to choose 2 out 3 available pilots 3C2 = 3
The only other possibility is that Hugo is not the pilot. That happens when one of the 3 other pilots are chose. Any 2 copiilots can be chosen as the other pilots have no constraint on who can be their co-pilots.
Thus, ways of choosing 1 pilot from 3 available pilots = 3C1 = 3
ways of choosing 2 co-pilots from 6 available co-pilots = 6C2 = 15
Thus, total combinations where Hugo is not the pilot = 3*15 = 45
Overall total number of combinations = combinations where Hugo is pilot (3) + Comibinations where he is not the pilot (45) = 48
Thus P (Hugo is pilot) = 3/48 = 1/16
Column 2: Exactly one of the two co pilots has experience > 500 hrs.
Since we only have to choose 2 co-pilots, one will have experience > 500 hrs and the other will not. We have 3 available in each category
Thus,
Ways of choosing a co pilot who has experience greater > 500 hrs = 3C1 = 3 and the same for the co pilot experience < 500 hrs = 3C1 = 3
Thus, total combinations of co pilots = 3*3 = 9
We will also have 3 pilots who can fly with these combinations of co pilots (Hugo cannot, since both are not experience > 500).
Thus, total number of combinations with 1 pilot and 2 co pilots, such that exactly one co pilot has experience > 500 hrs = 3*9 = 27
Hence, Probability = 27/48 = 9/16
