Problem 15 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least \(90 \%\) is : \(\quad\) [April. 08, 2019 (II)] (a) 5 (b) 3 (c) 4 (d) 2

Step-by-Step Solution

Answer

The minimum number of tosses is 4.

Step 1: Understand the Problem

We need to find the minimum number of times to toss a coin such that the probability of getting at least one head is at least 0.9 (or 90%).

Step 2: Express the Probability of Getting At Least One Head

The probability of getting at least one head in n coin tosses is the complement of getting all tails. Hence, we have: \[ P(\text{at least one head}) = 1 - P(\text{all tails}) \] where the probability of getting all tails in n tosses is \((0.5)^n\).

Step 3: Set Up the Probability Inequality

We need \( 1 - (0.5)^n \geq 0.9 \). This ensures the probability of observing at least one head is at least 90%.

Step 4: Solve the Inequality

Rearrange and solve the inequality: \[ 1 - (0.5)^n \geq 0.9 \] \[ (0.5)^n \leq 0.1 \]

Step 5: Test Values of n

Calculate \((0.5)^n\) for different values of n to find the smallest n that satisfies the inequality:- For \( n = 1 \), \( (0.5)^1 = 0.5 \)- For \( n = 2 \), \( (0.5)^2 = 0.25 \)- For \( n = 3 \), \( (0.5)^3 = 0.125 \)- For \( n = 4 \), \( (0.5)^4 = 0.0625 \)The smallest n for which \((0.5)^n \leq 0.1\) is \( n = 4 \).

Step 6: Verify the Solution

When n = 4, \( (0.5)^4 = 0.0625 \), which is less than or equal to 0.1, hence the requirement is satisfied. Therefore, we need at least 4 tosses.

Key Concepts

Complementary Events

In probability theory, complementary events are two mutually exclusive events that together cover all possible outcomes of an experiment. When considering the problem of finding the probability of tossing at least one head in multiple coin tosses, the concept of complementary events is a powerful tool.

When we say at least one head is tossed, it means avoiding the scenario where no heads (all tails) appear. The probability of getting all tails in a single toss of a fair coin is 0.5, and this probability is compounded when multiplying for multiple tosses.

Using complementary events simplifies calculations: instead of counting when at least one head appears, we calculate the probability of the complement (all tails) and subtract from 1. Thus, the probability of at least one head appearing in \( n \) tosses is:

• \( P(\text{at least one head}) = 1 - P(\text{all tails}) \)
• \( P(\text{all tails}) = (0.5)^n \)

This transformation avoids complex counting and provides direct ways to assess probabilities.

Inequalities

Inequalities allow us to determine ranges of solutions that satisfy certain conditions. In probability, these are often applied to find minimum or maximum occurrences of an event meeting a specified probability.

In the exercise, we need the probability of getting at least one head to be at least 90%. Setting up the inequality involves rearranging terms to isolate \( n \) on one side:

• Start with: \( 1 - (0.5)^n \geq 0.9 \)
• Simplify to: \( (0.5)^n \leq 0.1 \)

This form identifies the least number of tosses needed for the probability to reach or exceed 90%.

The inequality approach is crucial because it helps evaluate what values satisfy the required condition, turning the problem into a manageable calculation.

Exponential Functions

Exponential functions are central in probability calculations, especially in scenarios of repeated independent events like coin tosses. The phrase "\( (0.5)^n \)" represents an exponential function, where the base 0.5 (probability of tails per toss) is raised to the power of \( n \) (number of tosses).

This captures how probabilities change exponentially with each additional event. Exponential decay is evident here:

• As \( n \) increases, \( (0.5)^n \) gets very small quickly, approaching zero.
• For \( n = 1 \), probability is 0.5; for \( n = 2 \), it's 0.25; and it further reduces with more tosses.

Consequently, using exponential functions allows us to compute the likelihood of rare events when multiple independent trials occur, revealing how rapid shifts in probabilities occur with just a few additional trials.