We are asked to solve three probability problems: (b) A bag contains 54 red marbles and some blue marbles. 36% of the marbles in the bag are red. Find the number of blue marbles in the bag. (c) Another bag contains 15 red beads and 10 yellow beads. Ariana picks a bead at random, records its colour, and replaces it in the bag. She then picks another bead at random. (i) Find the probability that she picks two red beads. (ii) Find the probability that she does not pick two red beads. (d) A box contains 15 red pencils, 8 yellow pencils, and 2 green pencils. Two pencils are picked at random without replacement. Find the probability that at least one pencil is red.

1. Problem Description

We are asked to solve three probability problems:

(b) A bag contains 54 red marbles and some blue marbles. 36% of the marbles in the bag are red. Find the number of blue marbles in the bag.

(c) Another bag contains 15 red beads and 10 yellow beads. Ariana picks a bead at random, records its colour, and replaces it in the bag. She then picks another bead at random.

(i) Find the probability that she picks two red beads.

(ii) Find the probability that she does not pick two red beads.

(d) A box contains 15 red pencils, 8 yellow pencils, and 2 green pencils. Two pencils are picked at random without replacement. Find the probability that at least one pencil is red.

2. Solution Steps

(b)

Let $R$ be the number of red marbles and $B$ be the number of blue marbles.

We are given $R = 54$ and that 36% of the marbles are red.

Thus, $R = 0.36(R+B)$. Substituting $R = 54$, we have

$54 = 0.36(54+B)$.

Dividing by 0.36, we get

$54/0.36 = 54+B$

$150 = 54+B$

$B = 150 - 54 = 96$

(c)

(i) Let $R$ be the number of red beads and $Y$ be the number of yellow beads.

We have $R = 15$ and $Y = 10$. The total number of beads is $R+Y = 15+10 = 25$.

Since the bead is replaced after the first pick, the probability of picking a red bead on the first pick is $P(R_1) = \frac{15}{25}$.

The probability of picking a red bead on the second pick is $P(R_2) = \frac{15}{25}$.

The probability of picking two red beads is $P(R_1 \cap R_2) = P(R_1) \times P(R_2) = \frac{15}{25} \times \frac{15}{25} = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$.

(ii) The probability of not picking two red beads is the complement of picking two red beads.

$P(\text{not two red}) = 1 - P(\text{two red}) = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25}$.

(d)

Total number of pencils is $15+8+2 = 25$. We are picking two pencils without replacement.

The probability of at least one red pencil is 1 minus the probability of picking no red pencils (i.e., picking two non-red pencils).

The number of non-red pencils is $8+2 = 10$.

The probability of picking a non-red pencil on the first pick is $\frac{10}{25}$.

After picking a non-red pencil, there are 9 non-red pencils and 24 total pencils remaining.

The probability of picking another non-red pencil on the second pick is $\frac{9}{24}$.

The probability of picking two non-red pencils is $\frac{10}{25} \times \frac{9}{24} = \frac{2}{5} \times \frac{3}{8} = \frac{6}{40} = \frac{3}{20}$.

The probability of picking at least one red pencil is $1 - \frac{3}{20} = \frac{20}{20} - \frac{3}{20} = \frac{17}{20}$.

3. Final Answer

(b) The number of blue marbles is

9

6. (c)(i) The probability of picking two red beads is $\frac{9}{25}$.

(c)(ii) The probability of not picking two red beads is $\frac{16}{25}$.

(d) The probability that at least one pencil is red is $\frac{17}{20}$.