A coin is biased so that heads comes up thrice as likely as tails. In four independent tosses of the coin, what is probability of getting exactly three heads ?
Understanding the Probability Problem with a Biased Coin
The question asks for the probability of getting exactly three heads in four independent tosses of a coin that is biased. The bias means that heads is three times as likely to occur as tails.
Calculating Probabilities for a Biased Coin
Let \(P(H)\) be the probability of getting a head in a single toss, and \(P(T)\) be the probability of getting a tail in a single toss.
We are given that heads is thrice as likely as tails. This can be written as:
\(P(H) = 3 \times P(T)\)
Also, since heads and tails are the only possible outcomes, the sum of their probabilities must be 1:
\(P(H) + P(T) = 1\)
Now we can substitute the first equation into the second one to find \(P(T)\):
\(3 \times P(T) + P(T) = 1\)
\(4 \times P(T) = 1\)
\(P(T) = \frac{1}{4}\)
Now we can find \(P(H)\) using \(P(H) = 3 \times P(T)\):
\(P(H) = 3 \times \frac{1}{4} = \frac{3}{4}\)
So, the probability of getting a head in a single toss is \(\frac{3}{4}\), and the probability of getting a tail is \(\frac{1}{4}\).
Identifying the Binomial Distribution Context
This problem involves a fixed number of independent trials (four tosses), each with only two possible outcomes (heads or tails), and the probability of success (getting a head) is constant for each trial. This perfectly fits the description of a binomial probability problem.
We can define the parameters for the binomial distribution:
- Number of trials, \(n = 4\)
- Number of successes (heads) required, \(k = 3\)
- Probability of success (getting a head), \(p = P(H) = \frac{3}{4}\)
- Probability of failure (getting a tail), \(q = P(T) = 1 - p = 1 - \frac{3}{4} = \frac{1}{4}\)
Applying the Binomial Probability Formula
The probability of getting exactly \(k\) successes in \(n\) independent trials is given by the binomial probability formula:
\(P(X=k) = \binom{n}{k} p^k q^{n-k}\)
Where \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).
For our problem, we want the probability of exactly 3 heads in 4 tosses, so \(n=4\) and \(k=3\):
\(P(X=3) = \binom{4}{3} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^{4-3}\)
Step-by-Step Calculation
First, calculate the binomial coefficient \(\binom{4}{3}\):
\(\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times 1} = \frac{24}{6} = 4\)
Next, calculate the probability of getting 3 heads: \(\left(\frac{3}{4}\right)^3\)
\(\left(\frac{3}{4}\right)^3 = \frac{3^3}{4^3} = \frac{27}{64}\)
Then, calculate the probability of getting \(4-3=1\) tail: \(\left(\frac{1}{4}\right)^{4-3} = \left(\frac{1}{4}\right)^1\)
\(\left(\frac{1}{4}\right)^1 = \frac{1}{4}\)
Now, multiply these values together:
\(P(X=3) = 4 \times \frac{27}{64} \times \frac{1}{4}\)
\(P(X=3) = \frac{4 \times 27 \times 1}{64 \times 4}\)
\(P(X=3) = \frac{108}{256}\)
Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4:
\(P(X=3) = \frac{108 \div 4}{256 \div 4} = \frac{27}{64}\)
So, the probability of getting exactly three heads in four independent tosses of the biased coin is \(\frac{27}{64}\).
Summary of Probability Calculation
| Item | Value | Calculation/Reason |
|---|---|---|
| Bias Condition | \(P(H) = 3P(T)\) | Heads is thrice as likely as tails |
| Total Probability | \(P(H) + P(T) = 1\) | Sum of probabilities must be 1 |
| Probability of Tail | \(P(T) = \frac{1}{4}\) | Solving \(3P(T) + P(T) = 1\) |
| Probability of Head (p) | \(p = P(H) = \frac{3}{4}\) | \(3 \times P(T)\) |
| Probability of Tail (q) | \(q = \frac{1}{4}\) | \(1 - p\) |
| Number of Tosses (n) | \(n = 4\) | Given in the question |
| Number of Heads (k) | \(k = 3\) | Exactly three heads required |
| Binomial Coefficient | \(\binom{4}{3} = 4\) | Formula \(\frac{n!}{k!(n-k)!}\) |
| Binomial Probability | \(P(X=3) = \binom{4}{3} p^3 q^1\) | Applying formula |
| Final Probability | \(\frac{27}{64}\) | \(4 \times (\frac{3}{4})^3 \times (\frac{1}{4})^1 = 4 \times \frac{27}{64} \times \frac{1}{4}\) |
The calculated probability matches one of the given options.
Revision Table: Key Concepts in Biased Coin Probability
| Concept | Description | Relevance to Problem |
|---|---|---|
| Biased Coin | A coin where \(P(H) \neq P(T)\) | The coin has \(P(H) = \frac{3}{4}\) and \(P(T) = \frac{1}{4}\) |
| Independent Tosses | Outcome of one toss does not affect others | Allows multiplication of probabilities for sequences of outcomes |
| Binomial Distribution | Used for probability of k successes in n independent Bernoulli trials | Applies here as each toss is a Bernoulli trial with fixed probability of success (Head) |
| Probability of Success (p) | Probability of the desired outcome in one trial | \(p = P(H) = \frac{3}{4}\) |
| Probability of Failure (q) | Probability of the other outcome in one trial | \(q = P(T) = \frac{1}{4}\) |
| Binomial Coefficient \(\binom{n}{k}\) | Number of ways to choose k successes in n trials | \(\binom{4}{3}\) counts ways to get 3 heads in 4 tosses (HHTT, HTHT, HTTH, THHT, THTH, TTHH) - correction, it's permutations of 3 Hs and 1 T: HHHT, HHTH, HTHH, THHH. The calculation \(\binom{4}{3}=4\) is correct for the number of arrangements. |
Additional Information: Understanding Binomial Probability
The binomial probability formula is very useful for problems involving a series of independent trials, each with two possible outcomes. It requires knowing the total number of trials (\(n\)), the number of successful outcomes you're interested in (\(k\)), and the probability of a successful outcome in a single trial (\(p\)). The probability of failure in a single trial is then \(q = 1-p\).
The formula accounts for two things:
- The probability of getting exactly \(k\) successes and \(n-k\) failures in one specific order (e.g., HHH T). This probability is \(p^k q^{n-k}\).
- The number of different orders in which exactly \(k\) successes and \(n-k\) failures can occur. This is given by the binomial coefficient \(\binom{n}{k}\).
Multiplying these two parts gives the total probability of getting exactly \(k\) successes in \(n\) trials, regardless of the order.
Similar Questions
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Important Questions from Probability
- If one event is unaffected by the outcome of another event, the two events are said to be:
1. Dependent
2. Independent
3. Mutually exclusive
4. Mutually inclusive - Let $M$ be a randomly chosen non-empty subset of $S = \{1, 2, 3, \dots, 2026\}$.
Which of the following is the probability that the product of all the elements of $M$ is even? - A fair coin is tossed 6 times. The probability of getting at least 5 heads is _________________. (rounded off to two decimal places)
- Consider a random experiment of rolling a fair die. The probability of getting an odd number or a number less than 4 is ______ (Rounded off to two decimal places).
- In a null hypothesis significance test, the significance level $\alpha$ was set to 0.01. For which one or more of the following $p$ values would the null hypothesis be rejected?
