Can anyone please help with this probability question?

Question

sixteen men are demolishing an old factory when part of it collapsed, injuring three of them. Out of the sixteen men three were apprentices.

What is the probability that all three injured men were apprentices?

I have the answer but can't work out the calculations required.

hope someone can help. thanks

Answer 1

The answer requires you to make an assumption about the relative probability of an apprentice compared to a non-apprentice getting injured. If the chances are equal and independent, then the answer is simply 1/binomial(3,16) = 3!x13!/16!= 1/(7x5x16)

If the probability of an apprentice being in the wrong place at the wrong time is higher than a non-apprentice, then one can assume that the probability is somewhere between 1/560 and 1.

Answer 2

The propability of 'any' 3 got injured is 3/16.
The probability of the 1st apprentice got injured is 3/16, the 2nd is 2/15 and the 3rd is 1/14.

So the answer is 3/16*2/15*1/14=1/560

Answer 3

Brilliant answer, Circle. However you should answer 1> assuming all men are equally likely to be injured (3/16)*(2/15)*(1/14)
2> Then put a shortened version of Circle's answer.

Now the tricky bit understand the difference and try and apply in all your statistics work and
understand that professional mathematicians can make this mistake.
The other mistake they make is that the fractions aren't always written in stone. For instance. A baby has a 1/1000 of cot death, therefore the chance of 3 siblings dying of cot death is 1 in a billion. This falsely assumes that a second/third aren't more likely to have the same unexplained reason for dying.

Remember these 2 things in your exams and you will have some excellent second answers to
questions. ( Always state your assumptions)

Answer 4

Probability of first being an apprentice 1/16
Probability of second 1/15
Probability of third 1/14

The answer is 1/16 x 1/15 x 1/14 = 1/3360

It's the same odds as winning a lottery with 16 balls and needing 3 numbers

I was totally wrong. Who'd listed to me?

3/16 x 2/15 x 1/14 = 1/560 is the right answer. Defo.

Answer 5

Probability(3 apprentices) and Probability(3 injured)
= (3/16)x(3/16)
= 9/256

Answer 6

I think the answer is 1/560 = 0.001785714

Number of combinations of n things taken r at a time is nCr = n! / [(n-r)! X r!], where "!" is called factorial, such as 3!=3x2x1=6.

There are 16C3 ways to choose injured 3 people at random out of 16:

16C3 = = 16! / [13! X 3!] = 560

There is 3C3 = only one way to choose 3 injured apprentices out of 3 total apprentices.

Therefore the answer is 1/560

Answer 7

P( being all apprentices)
=16!/13!3!
=14*15*16/1*2*3=560
{there is only one set of three
apprentices}

the probability of all being apprentices
is 1 in 560

i hope that this helps

Answer 8

since 3/16 men were injured and only three from the group were apprentices, we can deduct that this type of probability is continous, without replacement. this means that the demoninator would decrease as the changes goes on (soz for my crap explanation) ill give and exmple e.g. 1/10x1/9 etc.
so, for this case, it would be 1/16x1/15x1/14=1/3360 (not sure that is correct, i did it at the top of my head.

Answer 9

go with bty - 1/16 * 1/15 * 1/14

Answer 10

Maybe draw a probability tree?