Question

1. a bag contains 40 balls, each of which is black or gold. feridun reaches into the bag and randomly removes two balls. each ball in the bag is equally likely to be removed. if the probability that two gold balls are removed is \\( \frac{5}{12} \\), how many of the 40 balls are gold?
2. an integer \\( n \\), with \\( 100 \leq n \leq 999 \\), is chosen at random. what is the probability that the sum of the digits of \\( n \\) is 24?
3. billy and crystal each have a bag of 9 balls. the balls in each bag are numbered from 1 to 9. billy and crystal each remove one ball from their own bag. let \\( b \\) be the sum of the numbers on the balls remaining in billy’s bag. let \\( c \\) be the sum of the numbers on the balls remaining in crystal’s bag. determine the probability that \\( b \\) and \\( c \\) differ by a multiple of 4.
4. oi-lam tosses three fair coins and removes all of the coins that come up heads. then she tosses the coins that remain, if any. determine the probability that she tosses exactly one head on the second toss.
5. two bags each contain 10 balls, labelled with the positive integers from 1 to 10. each ball is equally like to be chosen. pierre removes one ball from each bag. determine the probability that the product of the numbers on the two balls he chooses is divisible by 10.

Explanation:

Let's solve problem 7:

Step 1: Understand the problem

We have a bag with 40 balls, each black or gold. We remove 2 balls, and the probability of removing 2 gold balls is $\frac{5}{12}$. We need to find the number of gold balls, let's call the number of gold balls $g$.

The probability of choosing 2 gold balls from $g$ gold balls and 40 total balls is given by the combination formula. The number of ways to choose 2 gold balls from $g$ is $C(g, 2) = \frac{g(g - 1)}{2}$, and the number of ways to choose 2 balls from 40 is $C(40, 2) = \frac{40 \times 39}{2}$.

Step 2: Set up the probability equation

The probability of choosing 2 gold balls is $\frac{C(g, 2)}{C(40, 2)} = \frac{5}{12}$.

Substitute the combination formulas:
$$\frac{\frac{g(g - 1)}{2}}{\frac{40 \times 39}{2}} = \frac{5}{12}$$

The 2 in the numerator and denominator cancels out:
$$\frac{g(g - 1)}{40 \times 39} = \frac{5}{12}$$

Step 3: Solve for $g$

Cross - multiply:
$$12g(g - 1)=5\times40\times39$$
First, calculate the right - hand side: $5\times40\times39 = 200\times39=7800$

So the equation becomes:
$$12g^{2}-12g - 7800 = 0$$
Divide the entire equation by 12:
$$g^{2}-g - 650=0$$

We can solve this quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=- 1$, $c=-650$) using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

First, calculate the discriminant $\Delta=b^{2}-4ac=(-1)^{2}-4\times1\times(-650)=1 + 2600 = 2601$

Then $g=\frac{1\pm\sqrt{2601}}{2}=\frac{1\pm51}{2}$

We have two solutions:
$g=\frac{1 + 51}{2}=\frac{52}{2}=26$ and $g=\frac{1-51}{2}=\frac{-50}{2}=-25$

Since the number of gold balls cannot be negative, we take $g = 26$

Answer:

The number of gold balls is 26.