Problem 18 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

In order to get at least once a head with probability \(\geq 0.9\), the number of times a coin needs to be tossed is (a) 3 (b) 4 (c) 5 (d) None of these

Step-by-Step Solution

Answer

The number of times the coin must be tossed is 4.

Step 1: Understand the Problem

We need to determine the number of times a fair coin should be tossed to have a probability of at least 0.9 to get at least one head.

Step 2: Calculate Probability of Not Getting a Head in One Toss

The probability of getting a tail in one coin toss is 0.5 since the coin is fair.

Step 3: Determine Probability of Getting All Tails in Multiple Tosses

If we toss the coin \(n\) times and get all tails, the probability is \((0.5)^n\).

Step 4: Calculate Probability of Getting at Least One Head

The probability of getting at least one head is the complement of getting all tails: \[ P( ext{at least one head}) = 1 - (0.5)^n \]

Step 5: Set Up Inequality and Solve for n

We need this probability to be at least 0.9, therefore: \[ 1 - (0.5)^n \geq 0.9 \] Rearranging gives: \[ (0.5)^n \leq 0.1 \]

Step 6: Calculate Minimum n

Calculate \(n\) such that \((0.5)^n \leq 0.1\): Start with \(n = 1\): \((0.5)^1 = 0.5\) Try \(n = 2\): \((0.5)^2 = 0.25\) Try \(n = 3\): \((0.5)^3 = 0.125\) Try \(n = 4\): \((0.5)^4 = 0.0625\) Thus, we find that \((0.5)^4\) is less than or equal to 0.1.

Key Concepts

Complementary Probability

Complementary probability is a fundamental concept in probability theory. It refers to the idea of finding the probability of the complement of an event. The complement of an event consists of all possible outcomes that are not part of the event itself.
This means that to find the probability of an event not occurring, we subtract the probability of the event occurring from 1. Mathematically, if the probability of an event occurring is given by \( P(A) \), then the probability of it not occurring is \( 1 - P(A) \).

• For example, if you want to calculate the probability of getting at least one head when you toss a coin multiple times, you first find the probability of not getting any heads (which is all tails) and subtract it from 1.
• In our case, the complement is getting all tails, which is calculated as \((0.5)^n\) where \(n\) is the number of tosses.

Using complementary probability helps simplify complex scenarios into easier calculations, especially when dealing with binomial probabilities.

Fair Coin Probability

The concept of a 'fair coin' probability is integral to understanding probabilistic experiments involving a coin toss. A fair coin is one that has no bias, meaning it has an equal chance of landing on heads as it does on tails with each toss.
This probability can be denoted as 0.5 or 50% for each of the two possible outcomes—heads or tails.

• Every toss of a fair coin is an independent event. This means the outcome of one toss does not affect the outcome of another.
• For a single toss, the probability of getting either heads or tails remains constant at 0.5.

When tossing a fair coin multiple times, these properties make calculating outcomes straightforward, allowing us to multiply probabilities of singular events for combined outcomes. Understanding this concept is crucial because it sets the foundation for calculating more complex probabilities, such as in our exercise, where multiple tosses are examined to understand cumulative outcomes.

Binomial Probability

Binomial probability is a type of probability related to experiments that can have exactly two possible outcomes in each trial—for instance, success or failure. Coin tossing is a classic example where each toss results in either a head or a tail.

• Binomial probability distribution can predict the likelihood of a certain number of successes (e.g., getting heads) over a series of independent trials (e.g., all coin tosses).
• The probability of a specific outcome can be calculated by multiplying the probability of a success raised to the power of the number of successes, multiplied by the probability of failure raised to the power of the number of failures, and factoring in combinations if necessary.

The exercise involves determining the number of tosses needed to achieve a 90% probability of getting at least one head. The critical part here is using binomial probability to find the number of trials needed to reach that threshold. We use the property that the events are independent and multiply the probability of a single event occurring across all events.
By understanding binomial probability, we can solve a range of similar problems where independent trials and binary outcomes are involved.