We are given that a jar contains 3 pennies (worth 1 cent), 6 nickels (worth 5 cents), and 6 dimes (worth 10 cents). A child selects 2 coins randomly without replacement, and $X$ represents the amount in cents of the selected coins.

Step 1: Total Coins in the Jar

The total number of coins is:

$3 \text{ (pennies)} + 6 \text{ (nickels)} + 6 \text{ (dimes)} = 15 \text{ coins}.$

Step 2: Total Ways to Select 2 Coins

The number of ways to choose 2 coins from 15 coins is given by the combination formula $\binom{n}{r}$, where $n$ is the total number of items and $r$ is the number of items to choose:

$\binom{15}{2} = \frac{15 \times 14}{2} = 105 \text{ ways}.$

Step 3: Probability Calculation for $X = 10$

We are looking for the probability that the sum of the selected coins equals 10 cents. The possible combinations of coins that give a total of 10 cents are:

• A penny (1 cent) and a nickel (5 cents): Total = 1 + 5 = 6 cents (not 10).
• A penny (1 cent) and a dime (10 cents): Total = 1 + 10 = 11 cents (not 10).
• A nickel (5 cents) and a dime (10 cents): Total = 5 + 10 = 15 cents (not 10).
• Two dimes (10 + 10 = 20) (not 10).

There is no valid combination for $X = 10$ as the combinations fail.

Thus, the probability $X = 10$ is:

$P(X = 10) = 0$

Step 4: Probability Calculation for $X = 11$

We are now looking for the probability that the sum of the selected coins equals 11 cents. The possible combinations of coins that give a total of 11 cents are:

• A penny (1 cent) and a dime (10 cents): Total = 1 + 10 = 11 cents.

Now, we calculate the number of ways to select this combination:

• We have 3 pennies and 6 dimes, so the number of ways to select 1 penny and 1 dime is:

$3 \times 6 = 18 \text{ ways}.$

Thus, the probability $X = 11$ is:

$P(X = 11) = \frac{18}{105} \approx 0.171.$

Final Answers:

1. The probability $X = 10$ is $\boxed{0}$.
2. The probability $X = 11$ is $\boxed{0.171}$.

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Would you like further clarification or explanations on any part of this? Also, here are some related questions:

1. How do you calculate probabilities for other sums of coin selections?
2. What would happen if the child selected 3 coins instead of 2?
3. What is the probability of selecting 2 nickels?
4. How does the probability change if there are more pennies and fewer dimes?
5. What is the expected value of the sum of the coins selected?

Tip: When calculating probabilities in situations like this, it is helpful to first list all possible combinations of coins that result in the desired sum.