Probablity Question

Stanford

Is anybody here knowledgable in the area of numbers and probability, I'm sstruggling with a LCOL problem?

KaneToad

https://www.boards.ie/discussion/2058331273/probablity-question

The chances of me being any help to you are low.

But, hopefully my reply will entice in some others more knowledgeable on the topic!

zv2

https://www.boards.ie/discussion/2058331273/probablity-question

Why not post the problem? Better chance of an answer.

Stanford

Good idea, its a Leaving Cert Ordinary level question but something is not quite right

Three passengers who did not survive a plane crash are to be awarded bravery medals posthumously , Each person can only get one medal and all three medals are different,

1. In how many different ways can the medals be assigned
2.  If the three passengers were given the same medal would the number of outcomes be different from that in 1. above? Why/why not?

If the probability that a passenger chosen was in first class is  0.4

What is the probability that

 

1.   The passenger chosen first was in first class

 

2.   All three passengers chosen were in first class

 

3.   At least one passenger chosen was not in first class

Michael Collins

https://www.boards.ie/discussion/comment/121619821#Comment_121619821

The question is badly phrased. Did you copy it word-for-word? It's not completely clear if each person must get a different medal, or if there are 3 distinct medal types which can be given out multiple times.

Since Q2 suggests the same medal can be given out multiple times, so I'll go with that assumption.

Q1. Each person can be given one of three different medals, so there are 3 x 3 x 3 = 27 possible arrangements.

Q2. Yes, if they all must be given the same medal then we have 3 x 1 x 1 = 3 possible arrangements.

For the probabilities

Q1. Assuming any given passenger has 0.4 probability of being in first class, then the first such passenger chosen has this probability of being in first class i.e. 0.4.

Q2. There's really no way to answer this, as we are not told that the passengers are chosen independently of each other. And, in principle, if this were a real scenario, since the same person can't be chosen twice, if the first passenger chosen is in first class then there are fewer passengers left in first class so this should affect the probability of the next passenger being in first class. Not to mention that they all died, which might indicate there were in the same section of the plane!

Q3. Same problem as above.

If we ignore this, and assume each passenger is chosen independently (somehow), then

Q2. 0.4 * 0.4 * 0.4 = 0.064

Q3. Since either all passengers are in first class, or at least some are not, we have:

P(all in 1st class) + [P(exactly one not in first class) + P(exactly two not in first class) + P(exactly three not in first class)] = 1

=> P(all in 1st class) + [P(at least one not in first class)] = 1

=> P(at least one not in first class) = 1 - P(all in 1st class) = 1 - 0.4*0.4*0.4 = 1 - 0.064 = 0.936

M.T. Cranium

For clarity, call the three people A, B and C, and the medals X, Y and Z.

I get only 6 as the different possible arrangements (not 27), if person A gets medal X, then two ways for persons B and C to get medals Y and Z, and if person A gets medal Y, also two ways, and if they get medal Z, also two ways, total of six arrangements.

It is basically same question as this: you take 3 kids out for ice cream, at the shop they only have one portion left of vanilla, chocolate and strawberry, so how many different ways are there to picture the outcome? You'll find it's six.

I agree with reasoning for other answers, if we are talking about a large number of people deceased; if the number is small, it would greatly affect the 0.4 probability of subsequent selections. In fact if the number of passengers on the plane is five and two are first class (and all perish) then the chance of at least one of three being "not first class" is 100%. If the number is ten, you get a value considerably higher than 93.6%, like 96.7% (and you need an infinite number of passengers to get exactly 93.6% but I'm sure it's close, under 94%, for say 100 passengers. It would be 1.0 minus (0.4 x 0.39 x 0.38).