The problem consists of two independent probability questions: (c)(i) A bag contains 15 red beads and 10 yellow beads. Ariana picks a bead at random, records its color, and replaces it in the bag. She then picks another bead at random. Find the probability that she picks two red beads. (c)(ii) Using the same scenario as (c)(i), 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

The problem consists of two independent probability questions:

(c)(i) A bag contains 15 red beads and 10 yellow beads. Ariana picks a bead at random, records its color, and replaces it in the bag. She then picks another bead at random. Find the probability that she picks two red beads.

(c)(ii) Using the same scenario as (c)(i), 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

(c)(i)

The total number of beads in the bag is $15 + 10 = 25$.

The probability of picking a red bead on the first pick is $\frac{15}{25} = \frac{3}{5}$.

Since the bead is replaced, the probability of picking a red bead on the second pick is also $\frac{15}{25} = \frac{3}{5}$.

The probability of picking two red beads is the product of the probabilities of picking a red bead on each pick:

$P(\text{two red beads}) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$.

(c)(ii)

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

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

(d)

Total number of pencils is $15 + 8 + 2 = 25$.

We want to find the probability that at least one pencil is red. It's easier to calculate the probability that no pencil is red, and subtract that from

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

The probability that the first pencil is not red is $\frac{10}{25} = \frac{2}{5}$.

If the first pencil is not red, there are now 9 non-red pencils and 24 total pencils.

The probability that the second pencil is not red, given the first one wasn't, is $\frac{9}{24} = \frac{3}{8}$.

The probability that neither pencil is red is $\frac{2}{5} \times \frac{3}{8} = \frac{6}{40} = \frac{3}{20}$.

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

3. Final Answer

(c)(i) $\frac{9}{25}$

(c)(ii) $\frac{16}{25}$

(d) $\frac{17}{20}$