Summary: (a) and (c)
Thank you for the question. I hope that I can help out here.
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First, let's talk about combinations.
The "choose" function gives the number of ways of choosing a subset of
a set of elements. That is the number of "combinations" of k from n
elements.
Specifically, choose(n,k) is the number of subsets of k items included
in a larger set of n items. choose(n,k) is usually written like this:
n
( )
k
But that is hard to read on a computer screen, so I will use
choose(n,k) .
In Matlab, choose(n,k) is called nchoosek(n,k). I definitely recommend
the Matlab software, http://www.mathworks.com , for exploring
statistics and many other things. I used Matlab for the computations
here, although actually any calculator, or even pencil and paper (do
students still use that?) would work.
For example, suppose n is 4 and k is 2. If the larger set has elements
{1,2,3,4}, then the possible 2element subsets are:
{1,2}
{1,3}
{1,4}
{2,3}
{2,4}
{3,4}
There are therefore 6 2element subsets of a 4element set, so
choose(4,2)=6.
The formula for computing choose(n,k) is n!/(k!(nk)!) where
n!=1*2*3*...*n.
Since 4!=4*3*2*1=24, and 2!=2, we see that choose(4,2)=
24/(2*2)=24/4=6, as we computed.
1. A binomial experiment is a little like flipping a coin, not
necessarily a fair coin. If we flip a coin that has a .06 chance of
reaching heads and a .94 chance of tails 7 times, what are the odds we
will get 2 heads?
Unfortunately the question is ambiguous. It is not clear whether the
questioner wants the probability of exactly 2 successes occuring or of
at least 2 heads occurring.
If the former, the probability is just
choose(7,2) * .06^2 * .94^5 .
If we imagine the coins flipped in order, there are
choose(7,2)
ways in which we can get exactly two heads. (That is, we can get heads
on say, 1st and second flip; or 1st and 3rd flip; or 2nd and 4th flip,
and so on).
For each such occurrence of heads, the probability it occurs is
.06^2*.94^5 . [For instance, the probability we get heads on first two
flips and tails on remaining 5 flips is this]
Now,
choose(7,2) = 7!/((5!*2!))=21
(you can verify this by just enumerating the subsets: {1,2}, {1,3},
and so on. You should get 21 subsets.
Hence, the probability we want is
21*.06^2*.94^5 = .055
rounded to two decimal points it's .06, so the answer is a.
So if the questioner meant "exactly two successes" the answer is a.
What if the questioner meant "at least two successes"? Then the
probability is the same as
1  P(zero or one successes)
Now, zero successes happens with P=.94^7.
1 success happens with
P=choose(7,1)*.06*.94^6 , so the probability we want is
1 (.94^7 + 7*.06*.94^6)= .0618 .
So this answer would be (a) also. Hence, no matter what the question
was supposed to mean, the correct answer is (a).
QUESTION 2

Very similar to the above problem.
Suppose he goes fishing M, T, W. (Monday, Tuesday, and Wednesday).
He can catch a fish on one day by either catching a fish M and not T
and not W; or T and not M and not W; or W and not M and not T.
Each of the three possibilities has P=.8*.2*.2 = .032 . Since there
are three mutually exclusive possibilities, the result is 3*.032 =
.096 . That is choose(3,1)*.032 .
Hence the answer is (c).
SEARCH STRATEGY

Well, I know all this stuff from years of professional experience, and
I am even more familiar with it since I just recently had to answer
another similar question about combinations here:
https://answers.google.com/answers/main?cmd=threadview&id=95478 . But
once you think of it the right way and do a lot of practice problems,
it's just common sense. 
Clarification of Answer by
rbnnga
on
03 Nov 2002 14:05 PST
The commenter is correct. I'd computed that the probability of
*exactly* two successes would be 0.0555 , which is close enough to the
requirement for (b), 0.0554 that indeed it is true that:
under the interpretation that *exactly* two successes are required,
the answer would be (b).
It's even MORE confusing though than this! That's because the two
options given are:
a) 0.06 and
b) 0.0554
Well, the trouble is that "0.06" IS 0.0554 rounded to two decimal
places. There is no way to know whether 0.06 is just a "rounding" of
0.0554  it depends on how much precision is known about the "0.06" as
the probability of success. In other words, the probability of success
only has one significant digit, arguably, so perhaps a) is just saying
we should not give the answer to greater precision than the question?
But why would the question in 1) be phrased WITHOUT the word "exact"
which WAS used in the same context in 2) ?
So, I guess there really is no way to know the "correct" answer here.
However, based on my experience with these kinds of homework
questions, I NOW think the intended answer is PROBABLY (b) for 1,
because:
i. a) is the same as P=0.06, so it could be a common mistake the book
is trying to trap you with.
ii. It is unlikely at this level that issues of precision would be
introduced, and
iii. It is quite possible that the word "exact" was just left off
the question statement
iv. The agreement is fairly close, and the "exact" interpretation is
easier mathematically than the "at least interpretation".
In conclusion, I am going with (b) now, agreeing with the commenter,
for question 1. Of course the math and derivation remain unchanged.
