Problem 5 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A coin is tossed \(n\) times. The probability of getting head at least once is greater than \(0.8\), then the least value of \(n\) is (a) 2 (b) 3 (c) 4 (d) 5

Step-by-Step Solution

Answer

The least value for n is 3.

Step 1: Understanding the Complement

To find the probability of getting at least one head, we first find the probability of getting no heads in any of the tosses. The outcome of getting no heads in a single toss is getting a tail, which has a probability of 0.5 for each toss.

Step 2: Calculate the Probability of No Heads

For each of the n tosses, getting tails is independent from one toss to another, so the probability of no heads (all tails) for n tosses is given by \( (0.5)^n \).

Step 3: Express the Desired Probability Condition

We want the probability of at least one head to be greater than 0.8. This is expressed as \( 1 - (0.5)^n > 0.8 \). This represents the complement rule, where the probability of at least one head is the complement of having no heads.

Step 4: Solve the Inequality

Starting with the inequality \( 1 - (0.5)^n > 0.8 \), we subtract 1 from both sides to get \( -(0.5)^n > -0.2 \). Multiplying by -1 (which flips the inequality), we have \( (0.5)^n < 0.2 \).

Step 5: Determine the Least Value of n

We solve for n by checking successive integer values starting from n=1. We find: \((0.5)^2 = 0.25\), \((0.5)^3 = 0.125\). When n=3, \( (0.5)^3 = 0.125 \), which is less than 0.2, satisfying \( (0.5)^n < 0.2 \).

Key Concepts

Binomial Distribution

The binomial distribution is a statistical distribution representing the number of successes in a series of independent trials, like flipping a coin. Each trial can end in either a success or a failure. In our coin toss example, getting a head can be considered a success. For the binomial distribution, several assumptions are made:

• There are a fixed number of trials, which is denoted by \( n \).
• Each trial is independent of the previous trials.
• The probability of success, in this case, getting a head, is constant at 0.5 for each toss.

The formula used to calculate these probabilities is \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( \binom{n}{k} \) is a combination, representing the different ways \( k \) successes can occur in \( n \) trials.

Complement Rule

The complement rule is a principle in probability that simplifies finding the probability of an event by considering the opposite event. To put it simply, the probability of an event happening is 1 minus the probability of it not happening. In our example of a coin toss, to find the probability of getting at least one head, which seems complex especially with many tosses, it's easier to calculate the probability of getting no heads at all.

• The probability of no heads (all tails) is \( (0.5)^n \), where each tail outcome has a probability of 0.5.
• This probability is then subtracted from 1 to find the probability of at least one head. So, \( P(\text{at least one head}) = 1 - P(\text{no heads}) \).

Using this rule can greatly ease the calculation, especially when dealing with a large number of trials, like coin flips.

Inequality Solving

Solving inequalities involves finding the range of values that satisfy a given condition. In our exercise, the inequality \( 1 - (0.5)^n > 0.8 \) describes the condition where the probability of getting at least one head is greater than 0.8. Here's how we go about solving it:

• First, we express the inequality in terms of \( (0.5)^n \). We begin by subtracting 1 on both sides to isolate the power term on the left: \( -(0.5)^n > -0.2 \).
• By multiplying both sides by -1, remembering to reverse the inequality sign, we get \( (0.5)^n < 0.2 \).
• To solve this, we test small values of \( n \) to find the smallest \( n \) that satisfies the inequality. We find that when \( n = 3 \), \((0.5)^3 = 0.125\) which is less than 0.2, thus satisfying the inequality.

Solving inequalities in probability problems often involves logical steps and testing small values.

Independent Events

In probability, independent events mean that the outcome of one event does not affect the outcome of another. This is a crucial concept when considering a series of trials such as coin tosses.

• Each coin toss in a series is considered an independent event because the result of one toss does not impact others.
• The probability of getting a head remains constant at 0.5 across all tosses.

Independence is essential when applying the binomial distribution, as it allows us to multiply probabilities of individual events to find the probability of a sequence of events, such as getting two tails in a row (\((0.5) \times (0.5) = 0.25 \)). Understanding independent events helps in approaching probability problems systematically and can simplify calculations significantly.