Question

\(A, B\) and \(C\) can solve \(50 \%, 60 \%\) and \(70 \%\) of the sums from a book. If one sum from that book is given them to solve then probability that the sum will be solved is (a) \(0.94\) (b) \(0.06\) (c) \(0.47\) (d) None

Short answer

The short answer is: The probability that the sum will be solved by at least one of them is \(0.94\).

Step-by-step solution

Write down the given probabilities of solving the problem

The three individuals have the probabilities of \(50 \%\), \(60 \%\) and \(70 \%\) respectively to solve the problem. Let's write it down: \(P(A) = 0.5\) \(P(B) = 0.6\) \(P(C) = 0.7\)

Calculate the probability of not solving the problem

Now we will calculate the probability that each individual will not solve the problem: \(P(A') = 1- P(A) = 1-0.5 = 0.5\) \(P(B') = 1- P(B) = 1-0.6 = 0.4\) \(P(C') = 1- P(C) = 1-0.7 = 0.3\)

Calculate the probability of all three not solving the problem

Now we need to find the probability that all three of them will not solve the problem. We will multiply the probabilities of the complementary events: \(P(A' \cap B' \cap C') = P(A') * P(B') * P(C') = 0.5 * 0.4 * 0.3 = 0.06\)

Calculate the probability of at least one solving the problem

Now, we will find the probability that at least one of them will solve the problem which is the complementary event we calculated in step 3: \(P(\text{at least one solves}) = 1 - P(A' \cap B' \cap C') = 1 - 0.06 = 0.94\)

Identify the answer

In the options provided, the correct answer is: (a) \(0.94\) So, the probability that the sum will be solved by at least one of them is \(0.94\).

Key concepts

Complementary Events

In probability theory, complementary events are a fundamental concept. A complementary event basically represents the opposite of a set event occurring. If the probability of an event happening is given by \( P(E) \), then the probability of the event not happening, or the complementary event, is \( P(E') = 1 - P(E) \).
\(E'\) is read as "E complement".
Consider the problem we have, where we want to solve a sum. If the probability that individual A can solve it is 0.5, then the probability that A cannot solve it is 0.5.

• This principle is crucial for solving many probability problems, especially when dealing with the sum of probabilities of overlapping events.
• In our case, this concept helped calculate the probability that none of the individuals solved the sum, which ultimately led to finding the probability that at least one did solve it.

Independent Events

Independent events are another key concept when it comes to probability. Two events are considered independent if the occurrence of one event does not influence the occurrence of the other. This means that the probability of both events happening together is simply the product of their individual probabilities.
In our original exercise, we had to compute the probability that A, B, and C could jointly fail to solve a problem. Since their attempts are independent, we could multiply their individual probabilities of failure: \( P(A' \cap B' \cap C') = P(A') \cdot P(B') \cdot P(C') \).

• This independence allowed us to simply use multiplication to derive the joint probability of all three failing to solve the sum.
• It emphasized how, for independent events, cumulative effects can be calculated easily, simplifying more complex probability calculations.

Probability Calculations

Probability calculations allow us to quantify uncertainty. By defining events and understanding relationships between them (like independence or complementarity), we can use simple arithmetic to calculate probabilities.
In our exercise, we used basic probability rules to methodically solve the problem step-by-step. We began by calculating probabilities for complementary events, then applied these to find joint probabilities for independent events.

• We began with individual probabilities: \(P(A), P(B), P(C)\).
• Then, calculated complementary probabilities: \(P(A'), P(B'), P(C')\).
• Multiplied complementary probabilities for a joint event: \(P(A' \cap B' \cap C')\).
• Finally, used complement principle to find "at least one" probability: \( 1 - P(A' \cap B' \cap C') \).

Learning how to break complex problems into simple, solvable tasks is a crucial part of mastering probability.

JEE Mathematics

JEE Mathematics often includes various types of probability problems, which challenge students to apply concepts of complementary events, independent events, and probability calculations. These questions assess both conceptual understanding and application skills.
The exercise solved here is typical of the kind of problems you might encounter in a JEE Mathematics exam wherein disciplined application of probability concepts is required.

• JEE problems often require understanding multiple layered concepts, urging a deeper comprehension beyond basic calculations.
• A solid grasp of probability theory can simplify solving complex JEE math problems.
• Consistently practicing these types of problems can develop the quick thinking skills necessary for success in the JEE examinations.

Grasping these concepts forms a foundation for further study in sciences and engineering, which are core elements of the Joint Entrance Examination syllabus.