Problem 14 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

In tossing 10 coins, the probability of getting exactly 5 heads is (a) \(9 / 128\) (b) \(63 / 256\) (c) \(1 / 2\) (d) \(193 / 256\)

Step-by-Step Solution

Answer

The probability of getting exactly 5 heads is (b) \(\frac{63}{256}\).

Step 1: Identify the Total Number of Outcomes

When tossing 10 coins, each coin has 2 possible outcomes: heads (H) or tails (T). So, the total number of possible outcomes when tossing 10 coins is \(2^{10} = 1024\).

Step 2: Recognize the Use of a Binomial Distribution

Each coin toss is a Bernoulli trial with two possible outcomes. To find the probability of getting exactly 5 heads in 10 tosses, we use the binomial probability formula where \(n = 10\) and \(k = 5\).

Step 3: Binomial Probability Formula

The binomial probability of exactly \(k\) successes (heads) in \(n\) trials (tosses) is computed as: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(p = 0.5\) is the probability of getting a head for each individual toss.

Step 4: Calculate the Binomial Coefficient

Compute the binomial coefficient \(\binom{10}{5}\) which represents the number of ways to choose 5 heads out of 10 coins: \[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = 252 \]

Step 5: Calculate the Probability Using the Coefficient

Substitute the coefficient and values into the binomial formula: \[ P(X = 5) = 252 \times (0.5)^5 \times (0.5)^5 = 252 \times (0.5)^{10} \]

Step 6: Compute the Probability

Calculate \((0.5)^{10} = \frac{1}{1024}\), then multiply by 252: \[ P(X = 5) = 252 \times \frac{1}{1024} = \frac{252}{1024} = \frac{63}{256} \].

Step 7: Conclude with the Correct Option

The probability of getting exactly 5 heads when tossing 10 coins is \(\frac{63}{256}\), which corresponds to option (b).

Key Concepts

Probability

Probability is the measure of the likelihood that an event will occur. When tossing a single coin, there are two possible outcomes: heads or tails. The probability of each outcome is equal, which is \(0.5\), or 50%. When it comes to multiple coin tosses, probability helps us determine the likelihood of a particular sequence of outcomes occurring.
This is crucial for problems involving repeated coin tosses, as it allows us to calculate the chance of getting a certain number of heads or tails.

• In a fair coin toss, each outcome (head or tail) has a probability of \(\frac{1}{2}\).
• The probabilities of all possible outcomes for any event should equal 1.

When extending this concept to 10 coin tosses, we look at all possible sequences. The total number of sequences (or outcomes) for 10 coins is \(2^{10} = 1024\). This foundational understanding helps us calculate the likelihood of specific results, such as getting exactly 5 heads in 10 tosses.

Binomial Coefficient

The binomial coefficient is a key component in the binomial probability formula and represents the number of ways to choose "successes" (like heads) from "trials" (coin tosses). It is mathematically represented as \(\binom{n}{k}\).
This coefficient is derived from combinations in mathematics, as opposed to permutations, since the order of selection does not matter.

• \(n\) signifies the total number of trials (e.g., 10 coin tosses).
• \(k\) is the number of successes we are interested in (e.g., 5 heads).

To compute the binomial coefficient, you can use the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
For our problem, \(\binom{10}{5}\) calculates to 252, indicating there are 252 distinct ways to get exactly 5 heads in 10 tosses. This step is essential to calculate the probability of specific events in binomial distributions.

Bernoulli Trial

A Bernoulli trial is an experiment or process that results in a binary outcome: success or failure. In the context of coin tossing, a Bernoulli trial occurs each time you flip a coin.
Each toss has two outcomes: heads (success) or tails (failure). Understanding this simplifies the process of calculating probabilities across multiple trials.

• Each trial is independent of the other, meaning the result of one toss doesn't affect the next.
• The probability of success remains constant throughout the trials.

When you toss a fair coin, the probability of landing on heads is \(p = 0.5\). When considering multiple Bernoulli trials, such as tossing a coin 10 times, you can use the binomial probability distribution to describe the outcome probabilities over the 10 trials.