From the image, you are asked to calculate the probability of picking a dime first and then a nickel from a set of coins without replacement. The set of coins is displayed on the right side of the question.

Step-by-step Solution:

1. Identify the coins:

Based on the image, the set consists of:

• 2 dimes
• 2 nickels
• 1 quarter

This makes a total of 5 coins.

2. First Event: Picking a dime.

The probability of selecting a dime first: $P(\text{dime}) = \frac{\text{Number of dimes}}{\text{Total coins}} = \frac{2}{5}$

3. Second Event: Picking a nickel.

Since the first coin is not replaced, one coin is now removed from the set. This leaves 4 coins. The number of nickels remains 2.

The probability of selecting a nickel given a dime was picked first: $P(\text{nickel | dime}) = \frac{\text{Number of nickels}}{\text{Remaining coins}} = \frac{2}{4} = \frac{1}{2}$

4. Combined Probability:

The combined probability of both events (dime first, then nickel) is: $P(\text{dime then nickel}) = P(\text{dime}) \times P(\text{nickel | dime}) = \frac{2}{5} \times \frac{1}{2} = \frac{2}{10} = \frac{1}{5}$

Final Answer:

$P(\text{dime then nickel}) = \frac{1}{5}$

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

Follow-Up Questions:

1. What is the probability of picking two nickels in a row?
2. How does the answer change if the first coin is replaced?
3. What is the probability of picking a quarter and then a dime?
4. If a nickel is picked first, what is the probability of picking another nickel?
5. How would this problem change if the total number of coins doubled?

Tip:

Always adjust the denominator of probabilities to reflect changes in the total number of coins when working without replacement.