Problem 15 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A coin is tossed three times in succession. If \(E\) is the event that there are at least 2 heads and \(F\) is the event in which the first throw is a head, then \(P(E / F)=\) (a) \(3 / 4\) (b) \(3 / 8\) (c) \(1 / 2\) (d) \(1 / 8\)

Step-by-Step Solution

Answer

The probability \( P(E | F) \) is \( \frac{3}{4} \).

Step 1: Understanding the problem

We are given a problem where a coin is tossed three times in succession. We identify event \( E \) as occurring if there are at least 2 heads and event \( F \) if the first toss is a head. We need to find the conditional probability \( P(E | F) \), the probability of having at least two heads given that the first toss is a head.

Step 2: Possible outcomes and event F

Since the first toss must be a head for event \( F \), our sample space for \( F \) based on the remaining tosses, consists of these outcomes: HHT, HHH, HTH, HTT. These sequences represent the outcomes when the first toss is fixed as a head.

Step 3: Count favorable outcomes for event E given F

We need at least 2 heads out of the three tosses. The outcomes satisfying this condition among our sample space for \( F \) are: HHT, HHH, and HTH. These are 3 favorable outcomes.

Step 4: Calculate the conditional probability

The conditional probability \( P(E | F) \) is given by the ratio of the number of favorable outcomes to the total number of possible outcomes given \( F \). Here that is \( \frac{3}{4} \) since there are 3 favorable outcomes out of 4 possible outcomes.

Key Concepts

Coin Toss

A coin toss is one of the purest forms of a probability experiment. When you flip a coin, there are two possible outcomes: heads (H) or tails (T). Each result in a single toss is equally likely, with a probability of 0.5 for getting either heads or tails. This makes coin tosses a common example in probability problems because of their simplicity and symmetry.
In our exercise, we are considering the tossing of a coin three times in a row. Repeated coin tosses are independent events, meaning the result of one toss doesn't affect the others. This independence is crucial because it allows us to calculate probabilities of sequences by multiplying the probabilities of individual outcomes.

Event Probability

Probability is the measure of the likelihood that a particular event will occur. For any event, the probability is calculated using the formula:

• Probability of Event = (Number of favorable outcomes) / (Total number of possible outcomes)

In our exercise, we have two specific events:

• Event E: At least 2 heads appear when tossed three times.
• Event F: The first toss results in a head.

The conditional probability question looks at the chance of event E happening given event F has occurred, denoted as \( P(E | F) \). To find this, we calculate how likely event E is within the subset of outcomes where event F is true.

Sample Space

The sample space in probability refers to the set of all possible outcomes of a probabilistic experiment. For the task at hand, since the first toss is fixed as a head due to event F, our sample space is limited to the results of the following two tosses.
Thus, the possible outcomes for three tosses given the first is a head (event F) would include: HHT, HHH, HTH, and HTT. Here, each string represents one way the tosses can play out. Altogether, there are 4 possible sequences in this new sample space. Recognizing the correct sample space is vital, as it directly influences our ability to correctly count favorable outcomes.

Favorable Outcomes

Favorable outcomes are the results that satisfy the criteria of an event. For analyzing probabilities, identifying these outcomes correctly is essential.
In our specific exercise, favorable outcomes are considered for event E within the sample space of event F. Event E demands at least two heads in three tosses. Given the earlier determined sample space of \{HHT, HHH, HTH, HTT\}, our favorable outcomes include:

• HHT: Two heads and one tail.
• HHH: Three heads.
• HTH: Two heads and one tail.

There are 3 favorable outcomes. Thus, the probability \( P(E | F) \) is calculated as 3 favorable outcomes divided by the 4 total possible outcomes, leading to the conclusion of 3/4.