To determine which mathematical sentence can be used to calculate the probability of selecting a student who wants the option to keep studying online or is a boy, we need to identify two probabilities:
- The probability of selecting a student who wants the option to keep studying online.
- The probability of selecting a boy.
Since there is overlap (some boys may also want to keep studying online), we need to apply the principle of inclusion-exclusion to avoid double-counting.
Let's define these probabilities:
- Let \( P(A) \) be the probability that a student wants to keep studying online.
- Let \( P(B) \) be the probability that a student is a boy.
- Let \( P(A \cap B) \) be the probability that a student is both a boy and wants to keep studying online.
From the table:
- The total number of students is \( 25 \).
- The number of students who want to keep studying online (both girls and boys) is \( 13 \).
- The number of boys is \( 10 \).
- The number of boys who want to keep studying online is \( 3 \).
Now we calculate the probabilities:
\[ P(A) = \frac{13}{25} \] \[ P(B) = \frac{10}{25} \] \[ P(A \cap B) = \frac{3}{25} \]
Using the inclusion-exclusion principle:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] \[ P(A \cup B) = \frac{13}{25} + \frac{10}{25} - \frac{3}{25} = \frac{13 + 10 - 3}{25} = \frac{20}{25} \]
This expression corresponds to the mathematical sentence:
\[ \frac{13}{25} + \frac{10}{25} - \frac{3}{25} \]
Therefore, the correct response is:
Start Fraction 13 over 25 End Fraction plus Start Fraction 10 over 25 End Fraction minus Start Fraction 3 over 25 End Fraction.