I'm going to assume that what you are asking is: if we roll a die until all sides (i.e. faces) appear, what is the probability that we will roll $n$ times?

In order for this to happen, we need exactly five faces to have appeared in the first $n-1$ rolls, and then the one missing face must appear on the $n$-th roll. Using inclusion-exclusion, we can express this, for $n>1$, as
$$
p(n) = \frac{1}{6} \binom{6}{5} \left( \left(\frac{5}{6} \right)^{n-1} - \binom{5}{4} \left( \frac{4}{6} \right)^{n-1} + \binom{5}{3} \left( \frac{3}{6}\right)^{n-1} - \binom{5}{2} \left( \frac{2}{6} \right)^{n-1}+\binom{5}{1} \left( \frac{1}{6}\right) ^{n-1} \right)
$$
So, for example, $p(6)=5/324$, $p(7)=25/648$, and $p$ hits a maximum of $875/10368$ at $n=11$.

As a check, you can (numerically) verify that
$$ \sum_{i=6}^{\infty} ip(i) = 14.7,$$
the well-known expected number of throws needed to see all faces.

(1) Your answer is correct. It appears you might be using software
instead of normal tables to get so many decimal places of accuracy.
[To use normal tables, you would have to 'standardize' (convert
to standard normal distributions), then get something like four digits of accuracy.] In R software, this computation is as follows, without standardizing.

```
qnorm(.25, 69.3, 2.8)
## 67.41143
1 - pnorm(67.4114, 64, 2.7)
## 0.1032081
```

In the graph below, 25% of the probability under the blue curve
(for men, at right) lies to the left of the dashed line at 67.4114,
and 10.32% of the probability under the orange curve lies to the
right of the same vertical line. (I recommend that you always try
to draw sketches for such problems, especially as problems become
more intricate than this one. Even very rough sketches can help
catch gross computational or logical errors.)

(2) Let $X$ be the height of a randomly chosen woman and $Y$
be the height of a randomly chosen man. This part requires you to look at the distribution of the difference $D = Y - X$.
Then $D$ is normally distributed with
$$E(D) = \mu_D = \mu_M - \mu_W = 69.3 - 64 = 5.3$$
and
$$V(D) = \sigma_D^2 = \sigma_M^2 + \sigma_W^2.$$
Notice that you subtract the means and add the $variances$.
(So far, you have been dealing with standard deviations.)
Then you want $P(D > 5.3).$ From what you have shown, I don't
think you should have trouble from there on. (Make a
sketch. Even without
computations, the answer should be obvious.)

I don't know if you care for simulations, but here are results
of a million simulated performances of this 2-person experiment.
Simulated results are not perfectly accurate, but you can use them as
a 'reality check' on your work.

```
x = rnorm(10^6, 64, 2.7)
y = rnorm(10^6, 69.3, 2.8)
d = y - x
mean(d); sd(d); mean(d > 5.3)
## 5.302365 # approx E(D)
## 3.889633 # approx SD(D)
## 0.50055 # approx P(D > 5.3)
```

(3) This part is very similar to part (2), but the result
is not obvious, and you have a little computation to do.
(My simulated answer is nearer to 0.53 than to 0.54.)

If this does not put you on the right track, or if you have
unresolved questions, please leave a Comment.

## Best Answer

$P(70)=P($living 5 more years$) \cap P(65)=(1-.16)(.746)=.62664.$

EDIT: Since we know that the probability that the man will live $5$ more years if he is $65$ is $1-.16=.84$. This is equivalent to saying:

$P(70|65)=.84$

Recall that $P(A|B)=\frac{P(A\cap B)}{P(B)}$. So $P(70|65)=\frac{P(70\cap(65)}{P(65)}=\frac{P(70)}{P(65)}$

$\therefore P(70)=P(65)P(70|65)$, yielding the above result.