, 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$.)