A problem is given to three students A, B and C, whose probabilities of solving the problem independently are \(\rm \frac{1}{2}\) , \(\rm \frac{3}{4}\) and p respectively. If the probability that the problem can be solved is \(\rm \frac{29}{32}\) , then what is the value of p?
Understanding the Probability Problem
The problem involves three students, A, B, and C, attempting to solve a problem independently. We are given the individual probabilities of students A and B solving the problem, and the overall probability that the problem can be solved by at least one of them. We need to find the probability that student C solves the problem, denoted by 'p'.
Defining Probabilities of Solving and Failing
Let \(P(A)\), \(P(B)\), and \(P(C)\) be the probabilities that students A, B, and C solve the problem, respectively. We are given:
- \(P(A) = \frac{1}{2}\)
- \(P(B) = \frac{3}{4}\)
- \(P(C) = p\)
Since the students work independently, the probability that a student *fails* to solve the problem is 1 minus the probability that they solve it.
- Probability that A fails: \(P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}\)
- Probability that B fails: \(P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}\)
- Probability that C fails: \(P(C') = 1 - P(C) = 1 - p\)
Using the Probability of the Problem Being Solved
The problem can be solved if at least one of the students solves it. This event is the complement of the event where *none* of the students solve the problem.
Let \(P(\text{Problem Solved})\) be the probability that the problem can be solved. We are given:
\(P(\text{Problem Solved}) = \frac{29}{32}\)
The probability that the problem is *not* solved is the probability that A fails AND B fails AND C fails. Since the events are independent, we can multiply their individual probabilities:
\(P(\text{Problem Not Solved}) = P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C')\)
\(P(\text{Problem Not Solved}) = \frac{1}{2} \times \frac{1}{4} \times (1 - p)\)
\(P(\text{Problem Not Solved}) = \frac{1}{8} (1 - p)\)
Relating Probabilities and Solving for p
The probability that the problem is solved is 1 minus the probability that it is not solved:
\(P(\text{Problem Solved}) = 1 - P(\text{Problem Not Solved})\)
Substitute the given value and the expression for \(P(\text{Problem Not Solved})\):
\(\frac{29}{32} = 1 - \frac{1}{8} (1 - p)\)
Now, we solve this equation for p:
Step 1: Isolate the term with p.
\(\frac{1}{8} (1 - p) = 1 - \frac{29}{32}\)
Step 2: Calculate the difference on the right side.
\(\frac{1}{8} (1 - p) = \frac{32}{32} - \frac{29}{32}\)
\(\frac{1}{8} (1 - p) = \frac{32 - 29}{32}\)
\(\frac{1}{8} (1 - p) = \frac{3}{32}\)
Step 3: Multiply both sides by 8 to isolate \((1 - p)\).
\(8 \times \frac{1}{8} (1 - p) = 8 \times \frac{3}{32}\)
\(1 - p = \frac{24}{32}\)
Step 4: Simplify the fraction.
\(1 - p = \frac{3}{4}\)
Step 5: Solve for p.
\(p = 1 - \frac{3}{4}\)
\(p = \frac{4}{4} - \frac{3}{4}\)
\(p = \frac{4 - 3}{4}\)
\(p = \frac{1}{4}\)
Thus, the value of p, the probability that student C solves the problem, is \(\frac{1}{4}\).
| Student | Probability of Solving | Probability of Failing |
|---|---|---|
| A | \(\frac{1}{2}\) | \(\frac{1}{2}\) |
| B | \(\frac{3}{4}\) | \(\frac{1}{4}\) |
| C | \(p\) | \(1-p\) |
Conclusion
The calculated value of p is \(\frac{1}{4}\), which means the probability that student C solves the problem is \(\frac{1}{4}\).
Revision Table: Key Concepts in Probability
| Concept | Description | Formula/Example |
|---|---|---|
| Probability of an Event E | The likelihood of event E occurring. Ranges from 0 to 1. | \(P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\) |
| Complement of an Event (E') | The event that E does NOT occur. | \(P(E') = 1 - P(E)\) |
| Independent Events | Events where the outcome of one does not affect the outcome of the others. | For independent events A and B, \(P(A \cap B) = P(A) \times P(B)\) |
| Probability of "At Least One" | The probability that at least one of several events occurs. Often calculated using the complement rule. | \(P(\text{At least one}) = 1 - P(\text{None})\) |
Additional Information: Independence in Probability
The concept of independent events is crucial in probability. When events are independent, the joint probability (the probability that all events occur) is simply the product of their individual probabilities. In this problem, the students solving the problem are independent events, meaning one student's success or failure does not influence another student's success or failure.
Understanding independence allows us to calculate the probability that none of the students solve the problem by multiplying their individual probabilities of failing:
\(P(\text{A fails AND B fails AND C fails}) = P(\text{A fails}) \times P(\text{B fails}) \times P(\text{C fails})\)
This relationship is key to solving problems where you are given the probability of an outcome occurring "at least once" across independent trials or events. By calculating the probability of the complementary event (none of the events occurring), you can easily find the desired probability.
The problem provided is a common type of question testing the understanding of independent events and the complement rule in probability.
Important Questions from Probability of Random Experiments
A, B, C and D are mutually exclusive and exhaustive events.
If 2P(A) = 3P(B) = 4P(C) = 5P(D), then what is 77P(A) equal to ?
A fair coin is tossed 6 times. What is the probability of getting a result in the 6t h toss which is different from those obtained in the first five tosses ?
Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is
A biased coin with the probability of getting head equal to \(\frac{1}{4}\) is tossed five times. What is the probability of getting tail in all the first four tosses followed by head ?
Three dice are thrown. What is the probability that each face shows only multiples of 3 ?
