probabilty that either Hania will be first in the row or..

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Guest
hi,
The eight members of the debating club pose for a yearbook photo. If they line up randomly, what is the probabilty that

a) either Hania will be first in the row or Aaron will be last?

I thought it would be something like 1/8(hania) + 1/8 (aaron)
but thats wrong, so how would I do this question?

b) Hania will be first and Aaron will not be last? I looked in the back for the answers, and thought it should be the same answer to a) but NOPEERRS.
Could you clarify why?

Thanks
 
Anna, you must learn the inclusion/exclusion rule and how to apply it.

\(\displaystyle P(H \cup A) + P(H) + P(A) - P(H \cap A)\).
The probability that Hanna is first or(union) Aaron is last equals the probability that Hanna is first plus the probability that Aaron is last minus the probability that Hanna is first and(intersection) Aaron is last. We subtract because we have counted some twice.

\(\displaystyle P(H) = P(A) = \frac{{7!}}{{8!}}\) put Hanna first there (7!) ways to put the ofhrt seven.
The same for putting Aaron last.
But in some of those cases we counted some cases twice. That when we put Hanna first and arrange the others Aaron came last.

\(\displaystyle P(H \cap A) = \frac{{6!}}{{8!}}.\)

Anna, now you show some work on part (b)!
 
Re: probabilty that either Hania will be first in the row or

Hello, Anna!

The eight members of the debating club pose for a yearbook photo.
Click to expand...

There are: \(\displaystyle \,8!\:=\:40,320\) possible arrangements.

If they line up randomly, what is the probabilty that:
a) either Hania will be first in the row or Aaron will be last?
Click to expand...

There are three cases to consider.

[1] Hania is first and Aaron is not last.

. . .We have: .H _ _ _ _ _ _ _
. . . . . . . . . . - . . . . . . . . . ↑
. . . . . . . . . . . . . . . . . .not Aaron

. . .There are 6 choices for the last person.
. . .The other six members can be arranged in \(\displaystyle 6!\) ways.
. . .Hence, there are: \(\displaystyle \L\,6\cdot6!\) arrangements.


[2] Aaron is last and Hania is not first.

. . .We have: ._ _ _ _ _ _ A
. . . - . . . . . . ↑
. . . . . . . . not Hania

. . .There are 6 choices for the first person.
. . .The other six members can be arranged in \(\displaystyle 6!\) ways.
. . .Hence, there are: \(\displaystyle \L\,6\cdot6!\) arrangements.


[3] Hania is first and Aaron is last.

. . .We have: .H _ _ _ _ _ _ A
. . .The other six members can be arranged in \(\displaystyle 6!\) ways.
. . .Hence, there are: \(\displaystyle \L\,6!\) arrangements.


So there are: \(\displaystyle \L\,6\cdot6!\,+\,6\cdot6!\,+\,6!\:=\:13\cdot6!\) ways that Hania is first or Aaron is last.


Therefore: \(\displaystyle \:p(\text{Hania first or Aaron last}) \:=\:\L\frac{13\cdot6!}{8!}\)\(\displaystyle \:=\:\L\frac{13}{56}\)


b) Hania will be first and Aaron will not be last?
Click to expand...

We worked out the numerator in case [1] above: \(\displaystyle 6\cdot6!\)

Therefore: \(\displaystyle P(\text{Hania first and Aaron not last}) \:=\:\L\frac{6\cdot6!}{8!}\)\(\displaystyle \:=\:\L\frac{3}{28}\)

 
Re: probabilty that either Hania will be first in the row or

anna said:
The eight members of the debating club pose for a yearbook photo. If they line up randomly, what is the probabilty that a) either Hania will be first in the row or Aaron will be last?
Click to expand...

Is that the inclusive OR or the exclusive OR?
 
umm thats its common, was that what you were asking?

I havnt learned bout exclusive or ineclusive

we learned mutually and non mutually
 
mutually- events with no intersection

and non mutually - events with intersection
 
In that case the answer you got from both of us is correct.
But note using "either/or" usually implies mutual (non-overlapping).
 
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