To solve this problem, let’s break it down step by step:

Given:

• Reyna has 5 coins worth 10 cents each and 4 coins worth 25 cents each.
• Total number of coins = $5 + 4 = 9$.
• She randomly selects two coins.
• We need the probability that the total value of the two coins is at least 35 cents.

Step 1: Total number of ways to choose 2 coins from 9

Using combinations: $\text{Total ways} = \binom{9}{2} = \frac{9 \times 8}{2} = 36$

Step 2: Favorable cases where the total value is at least 35 cents

To find the favorable outcomes, let’s analyze the possible pairs of coins:

Case 1: One 10-cent coin and one 25-cent coin

• Value = $10 + 25 = 35$ (meets the condition).
• Number of ways to choose one 10-cent coin = $\binom{5}{1} = 5$.
• Number of ways to choose one 25-cent coin = $\binom{4}{1} = 4$.
• Total ways for this case = $5 \times 4 = 20$.

Case 2: Two 25-cent coins

• Value = $25 + 25 = 50$ (meets the condition).
• Number of ways to choose two 25-cent coins = $\binom{4}{2} = \frac{4 \times 3}{2} = 6$.

Case 3: Two 10-cent coins

• Value = $10 + 10 = 20$ (does not meet the condition). Ignore this case.

Step 3: Total favorable outcomes

$\text{Favorable outcomes} = 20 \text{ (Case 1) } + 6 \text{ (Case 2) } = 26$

Step 4: Probability

$\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{26}{36} = \frac{13}{18}$

Final Answer:

The probability is $\frac{13}{18}$.

Let me know if you'd like further clarification or details! 😊

Related Questions:

1. How can you calculate probabilities involving more complex combinations of coins?
2. What if Reyna draws three coins? How would the problem change?
3. How does the total number of coins impact the probabilities of different outcomes?
4. Can you generalize this problem to coins of different values or numbers?
5. How does the concept of independence play a role in this type of probability problem?

Tip:

Always start by determining all possible outcomes and favorable outcomes carefully before calculating probabilities.