The probability that a person chosen at random is left handed (in hand writing) is 0.1 what is the probability that in a group of ten people there is one and only one who is left handed.

Answer

To solve the problem, we need to find the probability that in a group of 10 people, exactly one person is left-handed, given that the probability of a person being left-handed is 0.1.

### Step-by-Step Solution:

1. **Identify the probabilities**:
- Probability of being left-handed (p) = 0.1
- Probability of being right-handed (q) = 1 - p = 1 - 0.1 = 0.9

2. **Determine the number of trials and successes**:
- Number of trials (n) = 10 (the total number of people)
- Number of successes (k) = 1 (we want exactly one left-handed person)

3. **Use the Binomial Probability Formula**:
The probability of getting exactly k successes in n trials is given by the formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Where:
- \(\binom{n}{k}\) is the binomial coefficient, which calculates the number of ways to choose k successes from n trials.
- \(p^k\) is the probability of success raised to the number of successes.
- \(q^{n-k}\) is the probability of failure raised to the number of failures.

4. **Calculate the binomial coefficient**:
\[ \binom{10}{1} = \frac{10!}{1!(10-1)!} = \frac{10}{1} = 10 \]

5. **Calculate the probabilities**:
- \(p^k = (0.1)^1 = 0.1\)
- \(q^{n-k} = (0.9)^{10-1} = (0.9)^9\)

6. **Calculate \( (0.9)^9 \)**:
\[ (0.9)^9 \approx 0.387420489 \]

7. **Combine all parts to find the probability**:
\[ P(X = 1) = \binom{10}{1} \cdot (0.1)^1 \cdot (0.9)^9 \]
\[ P(X = 1) = 10 \cdot 0.1 \cdot 0.387420489 \approx 0.387420489 \]

8. **Final Answer**:
The probability that in a group of 10 people, there is exactly one left-handed person is approximately **0.387**.