, 2024

Of the $7$ balls in the hat,
there are $3$ balls that are
even-numbered (numbered $2$, $4$ and $6$) and so the probability that the first
ball drawn is even-numbered is $\frac37$.
There are $7$ possible choices
for the first ball, and since a drawn ball is neither replaced nor
returned to the hat, there are $6$
choices for the second ball, and thus $7\times6=42$ ways that the first two balls
may be drawn.

The sum of the numbers on the first two balls drawn is $5$ exactly when the numbers are $1$ and $4$, in some order, or $2$ and $3$, in some order.

Thus there are $4$ possible ways that
the first two balls drawn have a sum of $5$: $1$
and $4$, $4$ and $1$, $2$
and $3$, or $3$ and $2$.

The probability that the sum of the numbers on the first two balls drawn
is $5$ is $\frac{4}{42}=\frac{2}{21}$.
Suppose the probability that the sum of the numbers on the first
two balls drawn is greater than or equal to $6$ is $p$.

Then we let $\overline{p}$ equal the
probability that the sum of the numbers on the first two balls drawn is
not greater than or equal to $6$.

That is, $\overline{p}$ is equal to
the probability that the sum of the numbers on the first two balls drawn
is less than $6$, and so $p=1-\overline{p}$.

If sum of the numbers on the first two balls drawn is less than $6$, then this sum is either $5$, $4$
or $3$ (since two different balls are
drawn, the smallest possible sum is $1+2=3$).

From part (b), there are exactly $4$
ways that the first two balls drawn have a sum of $5$.

There are exactly $2$ ways in which
the sum is $4$: $1$ and $3$ or $3$ and $1$ ($2$
and $2$ is not possible since there
is only one $2$).

There are exactly $2$ ways in which
the sum is $3$: $1$ and $2$ or $2$ and $1$.

Therefore, of the $7\times6=42$ ways
that the first two balls may be drawn, there are$4+2+2=8$ ways that the sum is less than
$6$, and so $\overline{p}=\frac{8}{42}=\frac{4}{21}$.

Finally, the probability that the sum of the numbers on the first two
balls drawn is greater than or equal to $6$ is $p=1-\overline{p}=1-\frac{4}{21}=\frac{17}{21}$.

Note: We may have instead chosen to determine $p$ directly. That is, we may have
determined the probability that the sum of the numbers on the first two
balls drawn was $6$, $7$, $8$, $9$, $10$, $11$, $12$, or $13$ and then added each of these
probabilities together to determine $p$.

We chose to determine $\overline{p}$
since it required considering that the sum of the numbers on the first
two balls drawn was $3$, $4$ or $5$, and thus was less work than it would
be to determine $p$
directly.
The probability that the sum of the numbers on the first two
balls drawn is greater than or equal to $7$ is $q=\frac34$.

As in part (c), we similarly define $\overline{q}$ to be the probability that
the sum of the numbers on the first two balls is less than $7$, and thus $q=1-\overline{q}$, or $\frac34=1-\overline{q}$, and so $\overline{q}=\frac14$.

There are $8$ possible choices for
the first ball (since an eighth ball was added to the hat) and $7$ choices for the second ball, and thus
$8\times7=56$ ways that the first two
balls may be drawn.

Since $\overline{q}=\frac14=\frac{14}{56}$, then
there are $14$ ways that the sum of
the numbers on the first two balls drawn is less than $7$.

Without using the new gold ball, there are $12$ ways that the sum of the numbers on
the first two balls drawn can be less than $7$.

These are: $1+5$, $1+4$, $1+3$, $1+2$, $2+4$, $2+3$, and their reversals.

Thus, the new gold ball, numbered with the integer $k$, where $1\leq k\leq7$, must give $2$ additional ways to produce a sum that
is less than $7$.

If $k=5$, then the gold ball may be
paired with the ball numbered $1$
(drawn in either order) to give $2$
additional ways to produce a sum that is less than $7$.

Further, if $k=5$, the gold ball
cannot be paired with any other ball to give a sum that is less than
$7$, and so the correct value of
$k$ is $5$.

(You should confirm for yourself that if $k=6
$\text{}$ or }7$, there are no additional ways for the sum to be
less than $7$, and if $k=1,2,3, \text{ or }4$, then there are
more than $2$ additional ways for the
sum to be less than $7$.)