Answer: Expected number of tests = 5.013
Step-by-step explanation:
Define random variable X that marks the number of tests required for some certain
group. Observe that if the test is negative for all the people (which has probability 0.95), we make one and only one test. If some of the people is tested positive, we make ten additional tests for every person in that group separately. Hence, the expected number of tests will be for if they are all negative (1 test) and the case of at least one person testing positive (11 tests).
That Is,
E(X) = 1(0.95^10) + 11(1 - (0.95^10))
E(X) = 0.5987 + 4.414 = 5.013
I'll give you an example from topology that might help - even if you don't know topology, the distinction between the proof styles should be clear.
Proposition: Let
S
be a closed subset of a complete metric space (,)
(
E
,
d
)
. Then the metric space (,)
(
S
,
d
)
is complete.
Proof Outline: Cauchy sequences in (,)
(
S
,
d
)
converge in (,)
(
E
,
d
)
by completeness, and since (,)
(
S
,
d
)
is closed, convergent sequences of points in (,)
(
S
,
d
)
converge in (,)
(
S
,
d
)
, so any Cauchy sequence of points in (,)
(
S
,
d
)
must converge in (,)
(
S
,
d
)
.
Proof: Let ()
(
a
n
)
be a Cauchy sequence in (,)
(
S
,
d
)
. Then each ∈
a
n
∈
E
since ⊆
S
⊆
E
, so we may treat ()
(
a
n
)
as a sequence in (,)
(
E
,
d
)
. By completeness of (,)
(
E
,
d
)
, →
a
n
→
a
for some point ∈
a
∈
E
. Since
S
is closed,
S
contains all of its limit points, implying that any convergent sequence of points of
S
must converge to a point of
S
. This shows that ∈
a
∈
S
, and so we see that →∈
a
n
→
a
∈
S
. As ()
(
a
n
)
was arbitrary, we see that Cauchy sequences in (,)
(
S
,
d
)
converge in (,)
(
S
,
d
)
, which is what we wanted to show.
The main difference here is the level of detail in the proofs. In the outline, we left out most of the details that are intuitively clear, providing the main idea so that a reader could fill in the details for themselves. In the actual proof, we go through the trouble of providing the more subtle details to make the argument more rigorous - ideally, a reader of a more complete proof should not be left wondering about any gaps in logic.
(There is another type of proof called a formal proof, in which everything is derived from first principles using mathematical logic. This type of proof is entirely rigorous but almost always very lengthy, so we typically sacrifice some rigor in favor of clarity.)
As you learn more about a topic, your proofs typically begin to approach proof outlines, since things that may not have seemed obvious before become intuitive and clear. When you are first learning it is best to go through the detailed proof to make sure that you understand everything as well as you think you do, and only once you have mastered a subject do you allow yourself to omit obvious details that should be clear to someone who understands the subject on the same level as you.
Four consecutive integers... n, n+1, n+2, n+3.
The product of the 1st and 4th is four less than twice the 1st multiplied by the 4th.
n(n+3)=2n(n+3)-4 perform indicated multiplications...
n^2+3n=2n^2+6n-4 subtract n^2 from both sides
3n=n^2+6n-4 subtract 3n from both sides
n^2+3n-4=0 factor
n^2-n+4n-4=0
n(n-1)+4(n-1)=0
(n+4)(n-1)=0, and since n>0
n=1
So the four numbers are 1, 2, 3, 4
check...
1(4)=2(1)4-4
4=8-4
4=4
This is valid through the law of syllogism. If you swap lines 1 and 2, then you'll have this argument:
If I step on a beehive, then I will get stung.
If I get stung by a bee, then it will hurt.
Therefore, if I step on a beehive, then it will hurt
------------------
So it's like connecting a chain together. Point A (stepping on the hive), leads to point B (getting stung), which leads to point C (getting hurt). We can take a shortcut to bypass point B to jump from A to C in one step. Check out the attached image for a visual of what I'm referring to.
