A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times

Q: Wait, why can't I just use $P(T) = \frac{1}{2}$? It's a coin, right?

That is a classic trap! In JEE problems, never assume a coin is fair unless the problem explicitly says so. The phrase 'biased so that a head is twice as likely' is a direct instruction to change your probability model. If you use $\frac{1}{2}$, you are ignoring the core constraint of the problem.

Q: Why is the combination $\binom{3}{2}$ equal to 3?

Great question. The binomial coefficient $\binom{n}{r}$ counts the number of ways to arrange $r$ successes in $n$ trials. For 3 tosses, the ways to get 2 tails are: (T, T, H), (T, H, T), and (H, T, T). That is exactly 3 arrangements. The formula $\binom{3}{2} = \frac{3!}{2!1!} = 3$ handles this for you automatically.

Q: Is it possible to solve this without the binomial formula?

Absolutely! You could list all 8 possible outcomes (HHH, HHT, HTH, THH, TTH, THT, HTT, TTT), calculate the probability of each specific sequence (e.g., $P(TTH) = \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{27}$), and then add up the probabilities of all sequences that contain exactly two tails. However, the binomial formula is much faster and less prone to calculation errors, which is why it is a favorite tool for JEE aspirants.

Question

A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

Options:

(A) $\frac{2}{9}$

(B) $\frac{1}{9}$

(C) $\frac{2}{27}$

(D) $\frac{1}{27}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Probability is not just about numbers; it is the mathematical language of uncertainty. When we encounter a problem involving a 'biased coin,' we are essentially being asked to recalibrate our intuition.

The problem states that a head is twice as likely to occur as a tail. If we denote the probability of a tail as $P(T) = x$, then the probability of a head must be $P(H) = 2x$.

Since the sum of all possible outcomes in a single toss must equal one, we have the fundamental equation:

$$P(H) + P(T) = 1$$

Substituting our variables, we get $2x + x = 1$, which simplifies to $3x = 1$. Solving this, we find $x = \frac{1}{3}$.

Thus, the probability of getting a tail is $P(T) = \frac{1}{3}$, and the probability of getting a head is $P(H) = \frac{2}{3}$. This is the foundation upon which we build our entire solution.

### The Binomial Framework

Now that we have our individual probabilities, we move to the next phase: the three tosses. We are tossing this biased coin three times, which means we have $n=3$ independent trials.

We are looking for a specific outcome: exactly two tails and one head. This is a classic scenario for the **Binomial Distribution**.

The Binomial Distribution allows us to calculate the probability of getting exactly $r$ successes in $n$ independent trials, where each trial has a probability of success $p$ and a probability of failure $q$. In our case, let us define 'success' as getting a tail.

Therefore, $p = P(T) = \frac{1}{3}$ and $q = P(H) = \frac{2}{3}$. We want $r=2$ successes in $n=3$ trials. The formula is:

$$P(X=r) = \binom{n}{r} p^r q^{n-r}$$

### The Final Calculation

Let us plug our values into this formula:

$$P(2T, 1H) = \binom{3}{2} \left( \frac{1}{3} ight)^2 \left( \frac{2}{3} ight)^1$$

First, we evaluate the combination $\binom{3}{2}$. This represents the number of ways to arrange two tails and one head in three tosses:

$$\binom{3}{2} = \frac{3!}{2!1!} = 3$$

Next, we calculate the powers:

$$\left( \frac{1}{3} ight)^2 = \frac{1}{9} \quad 	ext{and} \quad \left( \frac{2}{3} ight)^1 = \frac{2}{3}$$

Now, we multiply these components together:

$$3 	imes \frac{1}{9} 	imes \frac{2}{3} = \frac{2}{9}$$

The final probability is **$\frac{2}{9}$**. This journey from a simple biased coin to a precise probability value demonstrates the power of structured thinking.

Animated Solution for Mathematics - Probability: A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is-

Select Answer:

Visualized Solution

Understanding the Biased Coin

Total Probability Axiom

Individual Probabilities

The Three Tosses

Binomial Probability Formula

Substituting Values

Evaluating Combinations

Evaluating Powers

Final Calculation

Final Answer

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Binomial Framework

The Final Calculation

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