A word consists of 7 letters : 4 consonants and 3 vowels. If three letters are chosen at random, then the probability that more than one vowel will be selected, is

Options

A. `13/35`
B. `17/42`
C. `7/35`
D. `13/42`

Answer

To solve the problem, we need to find the probability of selecting more than one vowel when choosing 3 letters from a word that consists of 4 consonants and 3 vowels.

### Step-by-Step Solution:

1. **Identify Total Outcomes**:
We need to find the total number of ways to choose 3 letters from the 7 letters (4 consonants + 3 vowels).
\[ \text{Total outcomes} = \binom{7}{3} \]

2. **Calculate Total Outcomes**:
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

3. **Identify Favorable Outcomes**:
We want to find the cases where more than one vowel is selected. This can happen in two scenarios:
- Selecting 2 vowels and 1 consonant.
- Selecting 3 vowels.

4. **Calculate Favorable Outcomes for Each Scenario**:
- **Case 1**: Selecting 2 vowels and 1 consonant.
\[ \text{Ways to choose 2 vowels from 3} = \binom{3}{2} = 3 \]
\[ \text{Ways to choose 1 consonant from 4} = \binom{4}{1} = 4 \]
Therefore, the total ways for this case:
\[ \text{Total for Case 1} = \binom{3}{2} \times \binom{4}{1} = 3 \times 4 = 12 \]

- **Case 2**: Selecting 3 vowels.
\[ \text{Ways to choose 3 vowels from 3} = \binom{3}{3} = 1 \]
Therefore, the total ways for this case:
\[ \text{Total for Case 2} = \binom{3}{3} \times \binom{4}{0} = 1 \times 1 = 1 \]

5. **Total Favorable Outcomes**:
Adding both cases together:
\[ \text{Total favorable outcomes} = 12 + 1 = 13 \]

6. **Calculate Probability**:
The probability of selecting more than one vowel is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{more than 1 vowel}) = \frac{\text{Total favorable outcomes}}{\text{Total outcomes}} = \frac{13}{35} \]

### Final Answer:
The probability that more than one vowel will be selected is \(\frac{13}{35}\).