Bob

]]>The question doesn't make clear if you are expected to fit a formula to this data, such as Poisson. Also it says nothing about the time scale under which the data was collected. If, for example, we had been told " 'x' shows the number of machine breakdowns during each month collected over the last several years" , we could then estimate the probability of getting more than 2 breakdowns in the next week.

As we're not told this and not asked for such answers, I'd take the simplest interpretation and just calculate total f = T and the probability that X > n by adding a third row to the table. Then for example P(X>6) = (4 + 3 + 1)/T

[X only takes discrete values so a continuous function is not appropriate I think]

Bob

If you have been given an example of a question like this, then post it and it may make things clearer.

]]>Similarly none of the terms is divisible by 3, for

It follows that

i.e. consecutive terms are coprime; also

i.e. alternate terms are coprime as well.

Since *a*₁ = 1, *k* =1 fits the bill. And since the sequence is clearly strictly increasing, it is the only one that does.

The number *n* of Sylow 17-subgroups divides 5 × 7 = 35 and is ≡ 1 (mod 17). Possibilities are 1 and 35.

Suppose 35. Then the intersection of all these subgroups (any two of which only have the identity in common since 17 is prime) would contain 1 + 35×(17−1) = 561 elements, i.e. 560 elements of order 17. Now, if *P* is a Sylow 17-subgroup and *Q* is a Sylow 5-subgroup, then *PQ* would be a subgroup of order |*P*||*Q*|/|*P*∩*Q*| = 85; moreover it would be cyclic, since *P* and *Q* are cyclic and their orders are coprime. But a cyclic subgroup of order 85 has *φ*(85) = 64 generators, i.e. 64 elements of order 85 – and we already have 560 elements of order 17.

Hence there can only be 1 Sylow 17-subgroup.

]]>H

*xy*:*x*hides*y*in public,

h*xy*:*x*has*y*,

M*x*:*x*is modest,*a*: Angus,*p*: private bits.

Then the premises are:

∀

*x*((M*x*∧h*xp*)→H*xp*),

∀*y*(h*ay*→¬H*ay*).

The first premise is equivalent to

•∀*x*(¬H*xp*→¬(M*x*∧h*xp*)) ≡ ∀*x*(¬H*xp*→(¬M*x*∨¬h*xp*)).

Substituting *x* = *a* into it gives

•¬H*ap*→(¬M*a*∨¬h*ap*).

Substituing *y* = *p* into the second premise gives

•h*ap*→¬H*ap*.

It follows that

•h*ap*→(¬M*a*∨¬h*ap*) ≡ ¬h*ap*∨(¬M*a*∨¬h*ap*) ≡ ¬M*a*∨¬h*ap*

which is the conclusion. Therefore the argument is valid.

]]>Does it implies that the order of cyclic subgroup {1, r, r^2, ... , r^n-1} of D2n is n then?

Yes.

]]>- .

Then

and

so

is a partition of *A* since *f* is surjective and (as a bona fide function) is well defined. So, use the axiom of choice to pick for each *b* ∈ *B* exactly one *aᵇ* ∈ *f*⁻¹(*b*). This will give you an injection

- .