Question

In order to get atleast once a head with probability \(P \geq 0.9\), the number of times a coin needs to be tossed is : (a) 3 (b) 2 (c) 5 (d) 4

Short answer

Answer: 4

Step-by-step solution

Identify the probability of getting at least one head

To find the probability of getting at least one head, we can subtract the probability of not getting a head in all tosses from 1 (using the complement rule). The probability of not getting a head is simply the probability of getting a tail in each toss.

Express the problem in terms of the number of tosses

Let's denote the number of tosses by n. The probability of getting at least one head (P) can be expressed in terms of n as follows: P = 1 - (probability of getting tails in all n tosses). The probability of getting tails in each toss is 1/2, so the probability of getting tails in all n tosses is (1/2)^n.

Set up the inequality

Now, we want P to be greater than or equal to 0.9, so we can set up the inequality: 1 - (1/2)^n ≥ 0.9.

Solve the inequality for n

To solve the inequality, we can start by subtracting 0.9 from both sides: (1/2)^n ≤ 0.1. Now, we apply the logarithm to both sides to isolate n: n >= log_0.5(0.1).

Calculate the minimum value of n

Using a calculator, we find that log_0.5(0.1) is approximately 3.32. Since we cannot have a fractional number of tosses, we round up to the nearest whole number, which is 4.

Identify the correct answer

Based on our calculation, the minimum number of tosses needed to have a probability of getting at least one head greater than or equal to 0.9 is 4. So the correct answer is (d) 4.

Key concepts

Coin Toss

A coin toss is a simple yet classic example of probability. When you toss a fair coin, there are two possible outcomes: heads or tails. Each outcome is equally likely, with a probability of 0.5 or 50%. This makes coin tosses a favorite in probability exercises.

In problems like the one above, multiple coin tosses are considered together. The probability of getting a specific result, like at least one head, can be calculated using these basic principles. Understanding the outcomes of single and multiple coin tosses lays the foundation for more complex probability calculations.

Complement Rule

The complement rule is a fundamental concept in probability that simplifies calculations. If you want to find the probability of an event happening, sometimes it’s easier to find the probability of it not happening and subtract it from 1.

For example, if you want to find the probability of getting at least one head in multiple coin tosses, you can calculate the probability of getting no heads at all.

• The probability of getting a tail (not a head) in one toss is 0.5.
• The probability of getting only tails in all n tosses is \( \left( \frac{1}{2} \right)^n \).

By subtracting this from 1, you find the probability of getting at least one head. This approach can greatly simplify calculations in complex situations.

Exponential Function

In probability, exponential functions often appear in scenarios involving repeated independent events, like coin tosses. When calculating the probability of getting all tails in multiple tosses, you use an exponential function.For example, if each toss has a 0.5 chance of being tails, then the probability of getting tails n times in a row is \( \left( \frac{1}{2} \right)^n \). This exponential expression shows how quickly the probability decreases as the number of tosses increases.

Exponential functions help model situations where factors multiply repeatedly, making them crucial for understanding complex probability scenarios like the one in this exercise.

Logarithm

Logarithms are invaluable tools for solving equations in probability. When you have an exponential equation and need to find the exponent, logging both sides can isolate the variable.

In our exercise, we started with the inequality \( \left( \frac{1}{2} \right)^n \leq 0.1 \). To solve for n, we used the logarithm base 0.5 of 0.1, ensuring precision in calculations. Logarithms convert multiplication into addition, simplifying complex exponential expressions.

This makes them perfect for finding the number of trials required to achieve a certain probability, as seen in determining how many coin tosses are needed to likely get at least one head.