SOLUTION: 4. Three horses are running a race. Their names are Xeron (X), Yellowfly (Y), and Zonks (Z). They are equally likely to win the race.
(a) List all the possible orders in which the
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Question 1012223: 4. Three horses are running a race. Their names are Xeron (X), Yellowfly (Y), and Zonks (Z). They are equally likely to win the race.
(a) List all the possible orders in which these horses could finish the race.
(b) Mike bets that the horses will finish in the order YXZ. What is the probability that at least one of the horses will finish in the position that he predicts?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the horses are x,y,z.
the number of possible ordered arrangements is 3! = 3*2*1 = 6.
those arrangements are:
xyz
xzy
yxz
yzx
zxy
zyx
since there are 6 possible arrangements, the probability that any one of the arrangements will come true is 1/6.
that, however, doesn't appear to be what is being asked.
what is being asked is:
Mike bets that the horses will finish in the order YXZ. What is the probability that at least one of the horses will finish in the position that he predicts?
i think that means:
horse y will finish in position 1, or horse x will finish in position 2, or hoses z will finish in position 3.
there are 6 possible orders as reproduced below.
the chosen order is yxz.
the chosen position are therefore:
y is first, x is second, z is third, regardless of whether the chosen order was achieved or not.
here's the possible arrangements again, with the arrangements where at least one of the chosen positions was achieved.
xyz ***** z is third
xzy *****
yxz ***** y is first x is second z is third
yzx ***** y is first
zxy ***** x is second
zyx *****
it appears the probability that at least one of them will be in the chosen position is 4/6.
this shoud be supportive of the general formula where p(a or b or c) is equal to:
p(a) + p(b) + p(c) - p(ab) - p(ac) - p(bc) + p(abc).
let's see if that holds up.
event a is where y is first.
event b is where x is second.
event c is where z is third.
there are exactly 2 events each where these letters are in their chosen position.
therefore:
p(a) = 2/6
p(b) = 2/6
p(c) = 2/6
there are actually 3 events where ab or ac or bc are involved.
each of those events just happens to be the one event where all three are involved.
therefore:
p(ab) = 1/6
p(ac) = 1/6
p(bc) = 1/6
there is one event where all 3 are present.
therefore p(abc) = 1/6.
the formula of p(a or b or c) = p(a) + p(b) + p(c) - p(ab) - p(ac) - p(bc) + p(abc) becomes:
p(a or b or c) = 2/6 + 2/6 + 2/6 - 1/6 - 1/6 - 1/6 + 1/6 = 4/6.
the formula works, but it it wasn't all that easy to see why.
never the less, the probability should be accurate.
the probability that at least one of the horse will be in the chosen position based on the chosen order of yxz is 4/6.
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