Probability of skiing

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zzinfinity

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You really want to go on the annual ski trip with your construction company. There are 20 people on the project and 5 will be allowed to go. If the first three have already been chosen randomly, what is the probability that you'll be chosen to go? The decision is made through a random drawing from a hat.

I have two ideas for what the answer could be.

P(chosen on 4th draw) + P(chosen on 5th draw | not chosen on 4th draw)= 2/17+1/16

Or,
P(chosen on 4th draw) + P(chosen on 5th draw)= 2/17 + (1-2/17)*(1/16)

My reasoning on the second option is that you can only be chosen in the 5th draw if you are not chosen on the 4th draw so you need to account for that.

Does one of these seem more correct? Thanks!
 
This may be a stupid question, but are you supposed to assume that you were not chosen in the first three? Because wouldn't that then reduce this exercise to "100% less the probability of not being among the remaining two chosen"?
 
zzinfinity said:
You really want to go on the annual ski trip with your construction company. There are 20 people on the project and 5 will be allowed to go. If the first three have already been chosen randomly, what is the probability that you'll be chosen to go? The decision is made through a random drawing from a hat.

I have two ideas for what the answer could be.

P(chosen on 4th draw) + P(chosen on 5th draw | not chosen on 4th draw)= 2/17+1/16

Or,
P(chosen on 4th draw) + P(chosen on 5th draw)= 2/17 + (1-2/17)*(1/16)

My reasoning on the second option is that you can only be chosen in the 5th draw if you are not chosen on the 4th draw so you need to account for that.

Does one of these seem more correct? Thanks!
Click to expand...
For your first statement, use the intersection, not the conditional:

P(chosen on 4th draw) + P(chosen on 5th draw & not chosen on 4th draw) = . . .

which is exactly what you have calculated in your second statement! Thus

P(4th or (not4th & 5th)) = 2/17 + (1-2/17)*(1/16)
 
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