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The problem asks for the probability that both \(A\) and \(B\) will solve the problem, given certain odds.

### Given:
- The odds **against** \(A\) solving the problem are 10 to 7. 
  - This means the probability of \(A\) solving the problem, \(P(A)\), is:
    \[
    P(A) = \frac{7}{10 + 7} = \frac{7}{17}
    \]
- The odds **in favor** of \(B\) solving the problem are 15 to 12.
  - This means the probability of \(B\) solving the problem, \(P(B)\), is:
    \[
    P(B) = \frac{15}{15 + 12} = \frac{15}{27} = \frac{5}{9}
    \]

To find the probability that both \(A\) and \(B\) solve the problem, we assume their attempts are independent. The probability of both solving it is the product of their individual probabilities:
\[
P(\text{both A and B}) = P(A) \times P(B) = \frac{7}{17} \times \frac{5}{9} = \frac{35}{153}
\]

Thus, the probability that both \(A\) and \(B\) solve the problem is \(\frac{35}{153}\), which simplifies to \(\frac{5}{21}\).

Would you like further explanation or help with anything else?

Here are 5 related questions for further practice:
1. What are the odds against \(B\) solving the problem?
2. How do the odds of solving a problem relate to the probability of solving it?
3. If the odds in favor of \(C\) solving a problem are 4:5, what is \(P(C)\)?
4. What is the probability that neither \(A\) nor \(B\) solves the problem?
5. How would dependent events affect the calculation of probabilities?

**Tip:** When working with odds, remember that the total number of possible outcomes is the sum of the odds in favor and against.

The odds against A solving a problem are 10 to 7 and the odds in favour of B solving the problem are 15 to 12. What is the probability that if both of them try, the problem will be solved?

Learn how to calculate the probability that both A and B will solve a problem, given the odds against A solving it are 10 to 7, and the odds in favor of B solving it are 15 to 12. This solution involves key concepts in probability and uses the multiplication rule for independent events. Suitable for students in Grades 9-12.

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