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

Q: Wait, why is $X=3$ the maximum number of tosses? Can't the game go on forever?

Great observation! In many probability problems, games can indeed go on indefinitely. However, look closely at the stopping condition: 'until a head or three tails occur.' If you get three tails in a row, the game MUST stop. Therefore, it is impossible to have more than three tosses. The sample space is finite and small, which makes this problem elegant.

Q: Why did we include $TTT$ in the $X=3$ case? Doesn't the game stop at the third tail?

Exactly! The game stops at the third tail. If the sequence is $TTT$, the third toss is the one that triggers the 'three tails' condition. Thus, the game ends at $X=3$. It is a valid outcome for the random variable $X=3$.

Q: How do I know if I've accounted for all possible outcomes?

Check the sum of your probabilities: $P(X=1) + P(X=2) + P(X=3)$. If they sum to $1$, you have captured the entire sample space. In this case, $\frac{12}{16} + \frac{3}{16} + \frac{1}{16} = \frac{16}{16} = 1$. You are safe!

Q: Is there a shortcut using the formula for geometric distribution?

Be careful! The geometric distribution applies to an infinite series of Bernoulli trials until the first success. Because our game has a secondary stopping condition (three tails), it is not a standard geometric distribution. Stick to the tree method; it is more reliable for these 'truncated' scenarios.

Q: Why is the probability of Head $\frac{3}{4}$ and not $\frac{3}{3+1}$?

The problem states the head is 3 times as likely as the tail. If $P(T) = p$, then $P(H) = 3p$. Since the sum of probabilities must be $1$, we have $p + 3p = 1$, which gives $4p = 1$, so $p = \frac{1}{4}$. Your intuition is correct—it is indeed $\frac{3}{3+1}$!

Question

A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $X$ denotes the number of tosses of the coin, then the mean of $X$ is

Options:

(A) $\frac{38}{16}$

(B) $\frac{15}{16}$

(C) $\frac{21}{16}$

(D) $\frac{81}{64}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we aren't just solving a probability problem; we are mapping the life cycle of a game. Imagine you are holding a coin that is biased to favor the Head.

We are going to toss this coin, and the game will end either when we see the face of a Head or when we accumulate the misfortune of three Tails. Our goal is to find the **expected value** of the number of tosses, $X$.

### Decoding the Bias

Before we toss, we must understand our tool. The problem states that a Head is 3 times as likely as a Tail. Let the probability of a Tail be $p$. Then, the probability of a Head is $3p$.

Since these are the only two possible outcomes, their sum must be unity:
$$P(H) + P(T) = 1$$

Substituting our variables, we get $3p + p = 1$, which simplifies to $4p = 1$. Thus, we calibrate our instrument as follows:
$$P(T) = \frac{1}{4}, \quad P(H) = \frac{3}{4}$$

### Mapping the Stopping Conditions

Now, let us visualize the game. The random variable $X$ represents the number of tosses.

If the first toss is a Head, the game ends immediately ($X=1$). If the first is a Tail, we continue. If the second is a Head, the game ends ($X=2$).

If the second is also a Tail, we toss again. If the third is a Head, or if the third is a Tail (completing our set of three), the game ends ($X=3$). Thus, the possible values are $X \in \{1, 2, 3\}$.

### Calculating the Probabilities

For $X=1$, the sequence is simply $H$. The probability is:
$$P(X=1) = P(H) = \frac{3}{4}$$

For $X=2$, the sequence must be $TH$. Using the multiplication rule for independent events:
$$P(X=2) = P(T) \cdot P(H) = \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{16}$$

For $X=3$, the game reaches the third toss if we have already seen two Tails. The sequences that end at $X=3$ are $TTH$ and $TTT$. Both trigger the end of the game:
$$P(X=3) = P(TTH) + P(TTT)$$

Calculating these individual probabilities:
$$P(TTH) = \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{64}$$
$$P(TTT) = \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{64}$$

Summing these, we obtain:
$$P(X=3) = \frac{3}{64} + \frac{1}{64} = \frac{4}{64} = \frac{1}{16}$$

### The Expected Value

The mean, or expected value $E(X)$, is the weighted average of all possible outcomes:
$$E(X) = \sum x_i \cdot P(X=x_i)$$

Substituting our values:
$$E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3)$$

Plugging in the calculated probabilities:
$$E(X) = 1\left(\frac{3}{4}\right) + 2\left(\frac{3}{16}\right) + 3\left(\frac{1}{16}\right)$$

To add these, we use a common denominator of $16$:
$$E(X) = \frac{12}{16} + \frac{6}{16} + \frac{3}{16} = \frac{21}{16}$$

The final expected value of the number of tosses is **$\frac{21}{16}$**.

Animated Solution for Mathematics - Probability: A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $X$ denotes the number of tosses of the coin, then the mean of $X$ is

Select Answer:

Visualized Solution

Defining Probabilities

The Stopping Rule & Random Variable $X$

Case 1: Stopping at $X = 1$

Case 2: Stopping at $X = 2$

Calculating $P (X = 2)$

Case 3: Stopping at $X = 3$

Calculating $P (X = 3)$

Simplifying $P (X = 3)$

The Expected Value Formula

Substituting the Values

Making Denominators Common

Final Calculation

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

Decoding the Bias

Mapping the Stopping Conditions

Calculating the Probabilities

The Expected Value

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Comprehension Passage

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The probability that $X \geq 3$ equals

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Select Answer:

Visualized Solution

Defining Probabilities

The Stopping Rule & Random Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Case 1: Stopping at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=1

Case 2: Stopping at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=2

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=2)

Case 3: Stopping at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=3

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=3)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=3)