Solving a Probability Problem Involving Multiple Students

Consider a scenario where a problem is given to five students, and their respective chances of solving the problem are given as fractions. The aim is to calculate the probability that at least one student will solve the problem successfully. This problem can be effectively tackled using the principles of independent events and the complement rule.

Understanding the Problem

The problem states that the probabilities of solving the problem for each student are as follows:

Student 1: ( P_1 frac{1}{2} )

Student 2: ( P_2 frac{1}{3} )

Student 3: ( P_3 1 )

Student 4: ( P_4 frac{4}{5} )

Student 5: ( P_5 frac{2}{5} )

Student 6: ( P_6 frac{1}{6} )

Calculating the Probability of Each Student Not Solving the Problem

To find the probability that at least one student solves the problem, we can first calculate the probability that each student does not solve the problem:

( Ptext{not } S_1 1 - P_1 1 - frac{1}{2} frac{1}{2} ) ( Ptext{not } S_2 1 - P_2 1 - frac{1}{3} frac{2}{3} ) ( Ptext{not } S_3 1 - P_3 1 - 1 0 ) ( Ptext{not } S_4 1 - P_4 1 - frac{4}{5} frac{1}{5} ) ( Ptext{not } S_5 1 - P_5 1 - frac{2}{5} frac{3}{5} ) ( Ptext{not } S_6 1 - P_6 1 - frac{1}{6} frac{5}{6} )

Calculating the Probability That All Students Fail to Solve the Problem

Given that one of the students (the third one) has a probability of 1 of solving the problem, the probability that all students fail to solve the problem can be calculated by multiplying the probabilities of each student failing to solve the problem:

( Ptext{all fail} Ptext{not } S_1 times Ptext{not } S_2 times Ptext{not } S_4 times Ptext{not } S_5 times Ptext{not } S_6 )

( frac{1}{2} times frac{2}{3} times frac{1}{5} times frac{3}{5} times frac{5}{6} )

( frac{1 times 2 times 1 times 3 times 5}{2 times 3 times 5 times 5 times 6} )

( frac{30}{900} frac{1}{30} )

Calculating the Probability That at Least One Student Solves the Problem

The probability that at least one student solves the problem can be found by subtracting the probability that all students fail to solve the problem from 1:

( Ptext{at least one solves} 1 - Ptext{all fail} )

( 1 - frac{1}{30} )

( frac{29}{30} )

Thus, the probability that the problem will be solved is:

( frac{29}{30} ) or ( 0.9667 )

Using Complement Rule for Independence

Consider the probability that each student does not solve the problem:

( PA' 1 - frac{1}{2} frac{1}{2} ) ( PB' 1 - frac{1}{3} frac{2}{3} ) ( PC' 1 - frac{1}{4} frac{3}{4} )

The probability that none of them solves the problem can be calculated as:

( PA' times PB' times PC' frac{1}{2} times frac{2}{3} times frac{3}{4} frac{6}{24} frac{1}{4} )

The probability that at least one of the students solves the problem is the complement of none of them solving it:

( Ptext{at least one solves} 1 - frac{1}{4} frac{3}{4} )

Implications and Considerations

The mere disjunction of probabilities of each student solving the problem is not sufficient for the total probability calculation. Instead, we use the complement method to accurately determine the likelihood that at least one student solves the problem.

Understanding the complement rule and the independence of events in probability calculations is essential for solving such problems efficiently and accurately.