A candidate is selected for the interview of management trainees for 3 companies. for the first company therefore are 12 candidates for the second there are 15 candidates and fro the third, there are 10 candidates. find the probability that he is selected in at least one of the companies.

Answer

To find the probability that a candidate is selected in at least one of the three companies, we can use the concept of complementary probability. Here’s how we can solve the problem step by step:

### Step 1: Determine the total number of candidates for each company.
- For the first company, there are 12 candidates.
- For the second company, there are 15 candidates.
- For the third company, there are 10 candidates.

### Step 2: Calculate the probability of not being selected in each company.
- The probability of not being selected in the first company (P(A')) is:
\[ P(A') = 1 - P(A) = 1 - \frac{1}{12} = \frac{11}{12} \]

- The probability of not being selected in the second company (P(B')) is:
\[ P(B') = 1 - P(B) = 1 - \frac{1}{15} = \frac{14}{15} \]

- The probability of not being selected in the third company (P(C')) is:
\[ P(C') = 1 - P(C) = 1 - \frac{1}{10} = \frac{9}{10} \]

### Step 3: Calculate the probability of not being selected in any of the companies.
To find the probability of not being selected in any of the companies, we multiply the individual probabilities of not being selected:
\[ P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C') = \frac{11}{12} \times \frac{14}{15} \times \frac{9}{10} \]

### Step 4: Perform the multiplication.
Calculating the above expression:
\[ P(A' \cap B' \cap C') = \frac{11 \times 14 \times 9}{12 \times 15 \times 10} \]
Calculating the numerator:
\[ 11 \times 14 = 154 \]
\[ 154 \times 9 = 1386 \]
Calculating the denominator:
\[ 12 \times 15 = 180 \]
\[ 180 \times 10 = 1800 \]
Thus,
\[ P(A' \cap B' \cap C') = \frac{1386}{1800} \]

### Step 5: Calculate the probability of being selected in at least one company.
The probability of being selected in at least one company (P(A ∪ B ∪ C)) is given by:
\[ P(A ∪ B ∪ C) = 1 - P(A' \cap B' \cap C') = 1 - \frac{1386}{1800} \]
Calculating this gives:
\[ P(A ∪ B ∪ C) = \frac{1800 - 1386}{1800} = \frac{414}{1800} \]

### Step 6: Simplify the fraction.
To simplify \(\frac{414}{1800}\):
- The GCD of 414 and 1800 is 6.
\[ \frac{414 \div 6}{1800 \div 6} = \frac{69}{300} \]

### Final Answer:
Thus, the probability that the candidate is selected in at least one of the companies is:
\[ \frac{69}{300} \]