Five coins whose faces are marked 2, 3 are thrown. What is the probability of obtainining a total of 12 ?

Options

A. `1/16`
B. `3/16`
C. `5/16`
D. `7/16`

Answer

To solve the problem of finding the probability of obtaining a total of 12 when five coins marked with 2 and 3 are thrown, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have five coins, each showing either a 2 or a 3. We need to find the probability that the sum of the numbers shown on these coins equals 12.

2. **Identifying Possible Outcomes**:
Each coin has 2 faces (2 and 3). Therefore, the total number of outcomes when throwing 5 coins is given by:
\[ \text{Total Outcomes} = 2^5 = 32 \]

3. **Finding the Favorable Outcomes**:
To achieve a total of 12, we can denote the number of coins showing 2 as \( x \) and the number showing 3 as \( y \). Since there are 5 coins, we have:
\[ x + y = 5 \]
The total sum can be expressed as:
\[ 2x + 3y = 12 \]

4. **Setting Up the Equations**:
We can substitute \( y = 5 - x \) into the sum equation:
\[ 2x + 3(5 - x) = 12 \]
Simplifying this gives:
\[ 2x + 15 - 3x = 12 \implies -x + 15 = 12 \implies x = 3 \]
Thus, \( y = 5 - 3 = 2 \). This means we need 3 coins showing 2 and 2 coins showing 3.

5. **Calculating the Number of Arrangements**:
The number of ways to arrange 3 coins showing 2 and 2 coins showing 3 can be calculated using the combination formula:
\[ \text{Favorable Outcomes} = \frac{5!}{3! \cdot 2!} \]
Calculating this gives:
\[ = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{20}{2} = 10 \]

6. **Calculating the Probability**:
The probability of obtaining a total of 12 is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{Total} = 12) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{10}{32} = \frac{5}{16} \]

7. **Final Answer**:
Therefore, the probability of obtaining a total of 12 when throwing the five coins is:
\[ \frac{5}{16} \]

### Summary:
The probability of obtaining a total of 12 when five coins marked with 2 and 3 are thrown is \( \frac{5}{16} \).