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Beam me down
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I am doing some work on high school (Australia) level combinatorics. So I've been taught nCr, nPr, and of course factorial (!). Now I need to apply the combinatorics to probability.
There a two questions which I am stuck on.
I calcualted that the sample space of all combinations was [itex]2^8-1[/itex] since there has to be at least one ingredient for there to be a samwhich.
For part (a):
[tex]Pr(samwhich contains ham)=\frac {n(Ham)}{2^8-1}[/tex], where n(ham) is the number of combinations including ham.
The answer indicates that [itex]n(ham) = 2^7[/itex], but I don't know why it is [itex]2^7[/itex].
Part (b) is:
[tex]\frac {^8 C_3}{2^8-1}[/tex]
Part (c) I worked out to be (which is right according to the book):
[tex]\frac {\sum_{n=3}^8 ^8 C_n}{2^8-1}[/tex]
Another problem question for me is:
From that information I caculated that [itex]n(\epsilon) = ^18 P_3[/itex]. From that I caculated for (a) that the probaility of all red is:
[tex]Pr(all red) = \frac {5*4*3}{^18 P_3} = \frac {60}{4896} = \frac {5}{408}[/tex]
But the answer in the book is 5/204, so I'm off by a factor of 2, and I don't know why.
I calcualted part be to be:
[tex]Pr(all different colors) = \frac {7}{136}[/tex]
But the book says the answer is:
[tex]Pr(all different colors) = \frac {35}{136}[/tex]
There a two questions which I am stuck on.
Bill is making a sandwich. He may choose any combination of the following: lettuce, tomato, carrot, cheese, cucumber, beetroot, onion, ham. Find the probability that:
a. the sandwich contains ham
b. the sandwich contains three ingredients
c. the sandwich contains at least three ingredients
I calcualted that the sample space of all combinations was [itex]2^8-1[/itex] since there has to be at least one ingredient for there to be a samwhich.
For part (a):
[tex]Pr(samwhich contains ham)=\frac {n(Ham)}{2^8-1}[/tex], where n(ham) is the number of combinations including ham.
The answer indicates that [itex]n(ham) = 2^7[/itex], but I don't know why it is [itex]2^7[/itex].
Part (b) is:
[tex]\frac {^8 C_3}{2^8-1}[/tex]
Part (c) I worked out to be (which is right according to the book):
[tex]\frac {\sum_{n=3}^8 ^8 C_n}{2^8-1}[/tex]
Another problem question for me is:
A bag contains five white, six red and seven blue balls. If three balls are selected at
random, without replacement, find the probability they are:
(a) all red
(b) all different colours
From that information I caculated that [itex]n(\epsilon) = ^18 P_3[/itex]. From that I caculated for (a) that the probaility of all red is:
[tex]Pr(all red) = \frac {5*4*3}{^18 P_3} = \frac {60}{4896} = \frac {5}{408}[/tex]
But the answer in the book is 5/204, so I'm off by a factor of 2, and I don't know why.
I calcualted part be to be:
[tex]Pr(all different colors) = \frac {7}{136}[/tex]
But the book says the answer is:
[tex]Pr(all different colors) = \frac {35}{136}[/tex]
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