Problem 11 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(E\) and \(F\) are independent events such that 0 \(

Step-by-Step Solution

Answer

The correct answer is (d) All of the above.

Step 1: Understanding Independence of Events

Given two independent events, \(E\) and \(F\), independence means \(P(E \cap F) = P(E) \cdot P(F)\). First, let's verify whether \(E\) and \(F^c\) are independent.

Step 2: Checking Independence of E and F^c

Since \(E\) and \(F\) are independent, we check \(E\) and \(F^c\). We have \(P(F^c) = 1-P(F)\). Calculating \(P(E \cap F^c) = P(E) - P(E \cap F) = P(E) (1 - P(F))\), which simplifies to \(P(E) \cdot P(F^c)\), showing \(E\) and \(F^c\) are also independent.

Step 3: Checking Independence of E^c and F^c

Similarly, \(E^c\) and \(F^c\) need to check the independence condition: \(P(E^c \cap F^c) = P(E^c) \cdot P(F^c)\). Since \(E\) and \(F\) are independent, applying De Morgan's law, \(P(E^c \cap F^c) = 1 - P(E \cup F)\), which reduces to \((1-P(E))(1-P(F)) = P(E^c) \cdot P(F^c)\), thus \(E^c\) and \(F^c\) are independent.

Step 4: Verifying Conditional Probabilities Sum to 1

Check if \(P\left(\frac{E}{F}\right) + P\left(\frac{E^c}{F^c}\right) = 1\). This translates to \(\frac{P(E \cap F)}{P(F)} + \frac{P(E^c \cap F^c)}{P(F^c)} = P(E) + (1-P(F)) = P(E) + P(E^c) = 1\).

Step 5: Conclusion

All conditions listed in the options hold. Therefore, the correct choice is option (d) All of the above.

Key Concepts

Independent Events

In probability theory, two events are independent if the occurrence of one event does not influence or change the probability of the occurrence of the other event. This concept can be simply thought of as two events having no effect on each other. For instance, the fact that it rains today should not affect the probability of flipping a coin and getting heads or tails.

Mathematically, when two events, say \(E\) and \(F\), are independent, we express this relationship as:

• \(P(E \cap F) = P(E) \cdot P(F)\)

This means the probability of both \(E\) and \(F\) occurring together is just the product of their individual probabilities.

Moreover, when checking the independence of complementary events such as \(E\) and \(F^c\), we can also apply this rule. We already know \(P(F^c) = 1 - P(F)\). Hence, if the events \(E\) and \(F\) are independent, it follows that:

• \(P(E \cap F^c) = P(E) - P(E \cap F) = P(E) (1 - P(F)) = P(E) \cdot P(F^c)\)

So, \(E\) and \(F^c\) are also independent.

Complementary Events

Complementary events are a fundamental idea in probability. When discussing a single event \(F\), its complement, denoted by \(F^c\), consists of all outcomes in the sample space that are not part of \(F\). Simply put, if \(F\) represents some condition, \(F^c\) represents the opposite. For example, if event \(F\) is the event of rolling a dice and getting a number greater than 3, then \(F^c\) is rolling a number of 3 or less.

The probabilities of complementary events have a very neat and predictable relationship, which is:

• \(P(F) + P(F^c) = 1\)

This is because both events together cover the entire sample space and hence must sum to unity.

When analyzing independent events, it's important to consider their complements as well. In the case of events \(E\) and \(F\), if they are independent, then \(E\) is also independent of \(F^c\). Furthermore, \(E^c\) and \(F^c\) can also be checked for their independence using a similar approach:

• \(P(E^c \cap F^c) = (1 - P(E))(1 - P(F))\) indicates independence

Thus, even when events are flipped to their complements, the principle of independence remains consistent.

Conditional Probability

Conditional probability is a vital concept in understanding how the likelihood of one event depends on the occurrence of another. It is often phrased as "the probability of event \(E\) given event \(F\)". This is expressed with the notation \(P(E | F)\), and is calculated as:

• \(P(E | F) = \frac{P(E \cap F)}{P(F)}\)

This formula tells us how to update the probability of an event \(E\) once we know that another event \(F\) has occurred.

An interesting property of independent events is that the conditional probability \(P(E | F)\) is the same as the ordinary probability \(P(E)\), since:

• \(P(E | F) = \frac{P(E \cap F)}{P(F)} = \frac{P(E) \cdot P(F)}{P(F)} = P(E)\)

Moreover, when working with complementary events, we might encounter expressions like \(P\left(\frac{E}{F}\right) + P\left(\frac{E^c}{F^c}\right)\), and need to verify they sum to 1. This checks if the two conditional probabilities cover all possibilities and do so disjointly. Essentially, it ensures no overlap and complete coverage of the sample space in conditions. The result \(1\) signifies that if you account for both events and their complements, you account for the whole sample space.