Probability of A Solving Given

We are presented with a mathematics problem where three students, A, B, and C, attempt to solve it. The given information includes:

• The probability that student A successfully solves the problem, denoted as $P(A)$, is $\dfrac{1}{2}$.
• The probability that student B fails to solve the problem, denoted as $P(B')$, is $\dfrac{1}{4}$.
• The overall probability that the problem is solved by at least one of the students is $\dfrac{63}{64}$.

Our objective is to determine the probability that student C solves the problem, $P(C)$.

Probability Calculations for B and A

To solve this problem, we first derive probabilities that are directly related to the given information:

• The probability that student A does not solve the problem, $P(A')$, is calculated as: $$P(A') = 1 - P(A) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$$
• The probability that student B successfully solves the problem, $P(B)$, is calculated as: $$P(B) = 1 - P(B') = 1 - \dfrac{1}{4} = \dfrac{3}{4}$$

Problem Solved Probability Explained

The statement "the whole probability of the problem being solved is $\dfrac{63}{64}$" means that the probability of at least one student solving the problem is $\dfrac{63}{64}$.

We can use the concept of complementary events. The complement of the problem being solved is the event that *none* of the students solve the problem.

Let $P(\text{Solved})$ represent the probability that the problem is solved.

Let $P(\text{Not Solved})$ represent the probability that the problem is not solved by any student.

The relationship between these probabilities is:

$$P(\text{Solved}) + P(\text{Not Solved}) = 1$$

Using the given value, we find the probability that the problem is not solved:

$$P(\text{Not Solved}) = 1 - P(\text{Solved}) = 1 - \dfrac{63}{64} = \dfrac{1}{64}$$

Probability Calculation Using Independence

The event "Problem Not Solved" means that student A does not solve it, student B does not solve it, AND student C does not solve it. Assuming the individual efforts of the students are independent events, we can multiply their individual probabilities of not solving the problem:

$$P(\text{Not Solved}) = P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C')$$

We substitute the known probabilities into this equation:

• $P(\text{Not Solved}) = \dfrac{1}{64}$
• $P(A') = \dfrac{1}{2}$
• $P(B') = \dfrac{1}{4}$

The equation becomes:

$$\dfrac{1}{64} = \dfrac{1}{2} \times \dfrac{1}{4} \times P(C')$$

$$\dfrac{1}{64} = \dfrac{1}{8} \times P(C')$$

To find $P(C')$, the probability that student C does not solve the problem, we rearrange the equation:

$$P(C') = \dfrac{1}{64} \div \dfrac{1}{8}$$

$$P(C') = \dfrac{1}{64} \times 8$$

$$P(C') = \dfrac{8}{64}$$

$$P(C') = \dfrac{1}{8}$$

Probability of C Solving

Finally, we need to find the probability that student C solves the problem, $P(C)$. We use the relationship between solving and not solving:

$$P(C) = 1 - P(C')$$

Substituting the value of $P(C')$ we found:

$$P(C) = 1 - \dfrac{1}{8}$$

$$P(C) = \dfrac{8}{8} - \dfrac{1}{8}$$

$$P(C) = \dfrac{7}{8}$$

Thus, the probability of student C solving the mathematics problem is $\dfrac{7}{8}$.