1. A jar contains 20 red and 14 green marbles. Two marbles are selected, one at a time, without replacement.
ed?
b) What is the probability at least one of the marbles is red?

Let's break this problem down into two parts:

### Part a) What is the probability both marbles selected are red?

To find this, we'll use the basic probability rule:

\[
P(A \text{ and } B) = P(A) \times P(B \text{ given } A)
\]

Where:
- \(P(A)\) is the probability that the first marble selected is red.
- \(P(B \text{ given } A)\) is the probability that the second marble is red given that the first one was red (since we're selecting without replacement).

#### Step 1: Probability of selecting a red marble first
There are 20 red marbles and 34 total marbles (20 red + 14 green).

\[
P(\text{Red first}) = \frac{20}{34} = \frac{10}{17}
\]

#### Step 2: Probability of selecting a red marble second (after selecting a red first)
After selecting one red marble, there are 19 red marbles left, and only 33 marbles total remaining.

\[
P(\text{Red second given red first}) = \frac{19}{33}
\]

#### Step 3: Multiply the probabilities
\[
P(\text{Both red}) = \frac{10}{17} \times \frac{19}{33} = \frac{190}{561}
\]

So, the probability that both marbles selected are red is \( \frac{190}{561} \).

---

### Part b) What is the probability that at least one of the marbles is red?

We can approach this by first calculating the probability of the complementary event, which is the probability that **both marbles are green**, and then subtract this from 1:

\[
P(\text{At least one red}) = 1 - P(\text{Both green})
\]

#### Step 1: Probability of selecting a green marble first
There are 14 green marbles out of 34 total marbles.

\[
P(\text{Green first}) = \frac{14}{34} = \frac{7}{17}
\]

#### Step 2: Probability of selecting a green marble second (after selecting a green first)
After selecting one green marble, there are 13 green marbles left and 33 marbles total.

\[
P(\text{Green second given green first}) = \frac{13}{33}
\]

#### Step 3: Multiply the probabilities
\[
P(\text{Both green}) = \frac{7}{17} \times \frac{13}{33} = \frac{91}{561}
\]

#### Step 4: Use the complement rule
\[
P(\text{At least one red}) = 1 - \frac{91}{561} = \frac{470}{561}
\]

So, the probability that at least one of the marbles is red is \( \frac{470}{561} \).

---

### Summary:
a) The probability that both marbles are red is \( \frac{190}{561} \).

b) The probability that at least one of the marbles is red is \( \frac{470}{561} \).

Do you want more details or have any questions?

Here are five related questions you could explore:
1. What is the probability that both marbles are green?
2. What is the probability that the first marble is green and the second is red?
3. How does the probability change if the marbles are selected with replacement?
4. What is the expected number of red marbles in a sample of 2?
5. How would the probabilities change if there were 30 red and 14 green marbles?

**Tip:** The complement rule is a powerful tool in probability, especially when it's easier to calculate the complement of an event than the event itself.

A jar contains 20 red and 14 green marbles. Two marbles are selected, one at a time, without replacement. a) What is the probability that both marbles are red? b) What is the probability that at least one of the marbles is red?

Explore the probability of selecting red marbles from a jar with 20 red and 14 green marbles. Learn how to calculate the probability of both marbles being red and at least one being red using combinatorics and the complement rule. Suitable for students in Grades 8-10.

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