prob q hard! (1 Viewer)

[LostAuzzie]

I got 4/7 but I dont know if this is right:

Total number of ways the cards can be arranged:
n = 7!
= 5040
Ways in which first OR last are A:
n = 2 x 3 x 4 x 5!
= 2880
P(E) = 2880/5040
= 4/7

 

[klaw]

ok here's how i did it;

for A at start not at end

there are 7 spaces to be filled
for the 1st space it has to be 1 of the A's so 1/3
the last space can be filled by any of the remaining letters except the A's so 4/6
so total probability for A at start not at end is 1/4.4/6=2/9

you just double that for the other way; A at end not at start

hope thats right

why is it 1/3 and 4/6? isn't it 3/7 and 4/6?

 

[krabby_me]

the letters of the word BANANAS are written down, find the possibility when rearranged that A ill be first or last but not both.

1 - ((3/7)(4/6)

1- 30/42 = 1- 5/7 = 2/7

i think its wrong but its a go!

 

[LostAuzzie]

why is it 1/3 and 4/6? isn't it 3/7 and 4/6?

I agree because then it gives my answer

 

[speed2]

why is it 1/3 and 4/6? isn't it 3/7 and 4/6?

yeah ur right, silly me, i'll change it. den dat gives 4/7 but that ain't right so wtf?

 

[aaaman]

the letters of the word BANANAS are written on separate cards and drawn at random from a hat. calculate the probability that : an A will be drawn first or last but not both

ill give it a crack

P(meh) = (3/7 +3/7) - (1-(3/7*3/7))
= 2/49

 

[KFunk]

I got 4/7 but I dont know if this is right:

Total number of ways the cards can be arranged:
n = 7!
= 5040
Ways in which first OR last are A:
n = 2 x 3 x 4 x 5!
= 2880
P(E) = 2880/5040
= 4/7

That's the way I did it. I tried a few similar methods also which gave 4/7. So what is the correct answer then?

 

[krabby_me]

the letters of the word BANANAS are written down, find the possibility when rearranged that A ill be first or last but not both.



How about

there are 7! combinations of letters

the combinations of ways A can be implemented correctly are


AAABNNS
AABANNS
AABNANS
AABNNAS
ABANNAS
ABANANS
ABNAANS
ABNNAAS
ABNANAS

X that by the number of ways u arrange all the odd letters (4!/2) then by two to go from the other end. so u get

2x4!/2x9/ 7! = 2 x 24 x 9 / 5040
= 216/5040

 

[underthebridge]

I'm afraid there are not 7! arrangements to begin with -> you didn't account for repititions of A and N

 

[klaw]

We don't need to know the total number of arrangements

P(A drawn first or last)= P(A drawn first but not last)+P(A drawn last but not first)=3/7*2/3+4/7*1/2=4/7

P(A drawn first but not last):
1st:A (3/7)
Last: Anything but A (4/6)
2nd: Anything (5/5)
3rd: Anything (4/4)
4th: Anything (3/3)
5th: Anything (2/2)
6th: Anything (1/1)

P(A drawn last but not first)
1st:Anything but A (4/7)
Last: A (3/6)
2nd: Anything (5/5)
3rd: Anything (4/4)
4th: Anything (3/3)
5th: Anything (2/2)
6th: Anything (1/1)

P(A drawn first or last)= P(A drawn first but not last)+P(A drawn last but not first)=3/7*2/3+4/7*1/2=4/7

I don't see what's wrong

 

[KFunk]

I'm afraid there are not 7! arrangements to begin with -> you didn't account for repititions of A and N

It depends on how you work the question. The repeated arrangments of double/triple letters need to be factored in for probability because the repetition of an identical arrangment increases the chance that it will occur. But, as I said, it depends on how you go about calculating it.

 

[robbo_145]

AAABNNS
AABANNS
AABNANS
AABNNAS
ABANNAS
ABANANS
ABNAANS
ABNNAAS
ABNANAS

you missed one
ABAANNS

i would suggest from krabby's working that the answer is 4/7
10x2 placements of A (for mirroring)
3! ways of arranging A
4! ways of arranging the other letters
2x10x3!x4! = 2880
total combinations = 7! = 5040
P(E) = 2880/5040 = 4/7

 

[haboozin]

guys clam down....

<b>answer is 2/7</b>


Speed2 got it first i think..

 

[robbo_145]

guys clam down....

<b>answer is 2/7</b>


Speed2 got it first i think..

he realised he was wrong and changed it............do u actually have the solution or just the answer?

 

[KFunk]

Which book/test is it from?

 

[haboozin]

just an answer,

i donno where from, my friend just gave me this probability question that she couldn't do and i sucked at it n this guy posted it up..

 

[gemblack88]

Ummm... i dont know if im too late, and the answers already been worked out or whatever, but im gonna have a shot ne way. if its already been worked out or someones already tried this way and been proven wrong or i basically make a fool of myself in ne way (which i often seem to do on this thing... :/ ) then please ignore this!

Ok so here goes. If you take all the possible arrangements of the letters in BANANAS with A as the first letter like that other guy did (i didn't check it, i just counted it and he got ten, iincluding the one the girl added on so im just assuming thats right.) and then you multiply that by two (for all possible arrangements with A as the last letter) and then multiply THAT by 3! (for all possible arrangements of the three As) and then divide that by 420 (which is all of the possible arrangements of the letters in the word BANANAS: 7!/3!2!) well you then get 2/7, which thingo said was the answer so yeh. does that sound right?

(10 x 2 x 3!) / (7!/3!2!)

 

[robbo_145]

i think people are getting confused between the number of 'unique' arrangements that questions involving words usually look for (ie. most past hsc questions)
and probability ie number of combinations over total arrangements

 

[currysauce]

- hence Perms/Comb suck.