An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value k when k consecu

Q: Wait, why isn't $P(X=3)$ just $\binom{5}{3} / 32$?

That formula counts *any* 3 heads, not necessarily *consecutive* ones. The problem specifically demands consecutive heads, which restricts the valid arrangements significantly. We must count only the sequences where the heads appear in a block of exactly 3.

Q: Why do we subtract the probabilities for $X=3, 4, 5$ to get $P(X=-1)$?

Because the events $X=3, X=4, X=5,$ and $X=-1$ are mutually exclusive and collectively exhaustive. They cover every single possibility in the sample space, so their probabilities must sum to 1.

Q: Is there a faster way to count the sequences for $X=3$?

You can use a systematic approach by placing the block of 3 heads in different positions (start, middle, end) and ensuring the adjacent positions are tails to avoid creating a 4-head or 5-head sequence. For example, if the 3 heads are at the start, the 4th toss must be a tail.

Question

An unbiased coin is tossed 5 times. Suppose that a variable $X$ is assigned the value $k$ when $k$ consecutive heads are obtained for $k = 3, 4, 5$, otherwise $X$ takes the value $-1$. The expected value of $X$, is

Options:

(A) $\frac{1}{8}$

(B) $\frac{3}{16}$

(C) $-\frac{1}{8}$

(D) $-\frac{3}{16}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast probability space. We have five coins, and we are about to toss them into the air.

With each toss, the universe splits into two possibilities: heads or tails. Over five tosses, this creates a total sample space of $2^5 = 32$ equally likely outcomes. This is our foundation, the denominator for every probability we will calculate today.

### The "Exactly" Trap

Now, let's define our random variable $X$. The problem introduces a condition: $X$ takes the value $k$ if we get **exactly** $k$ consecutive heads, where $k \in \{3, 4, 5\}$.

If we fail to get at least three consecutive heads, $X$ takes a penalty of $-1$. The word "exactly" is our North Star here; it prevents us from overcounting.

For example, if we have $HHHHH$, that is $X=5$. If we have $HHHHT$, that is $X=4$. If we have $HHHTH$, that is $X=3$. We must be precise.

### Case-by-Case Analysis

For $X=5$, there is only one way: $HHHHH$. Thus, $P(X=5) = \frac{1}{32}$.

For $X=4$, we need exactly four consecutive heads. The sequence could be $HHHHT$ or $THHHH$. That gives us two favorable outcomes. Thus, $P(X=4) = \frac{2}{32}$.

Now, for $X=3$, we need exactly three consecutive heads. Let's place the block of three heads systematically:

1. At the start: $HHHTH$ and $HHHTT$. (Note: $HHHTH$ is valid because the 4th is a tail, and $HHHTT$ is valid because the 4th is a tail).

2. In the middle: $THHHT$. (The 1st and 5th must be tails to keep it exactly 3 heads).

3. At the end: $HTHHH$ and $TTHHH$. (The 2nd must be a tail to keep it exactly 3 heads).

That gives us five favorable outcomes. Thus, $P(X=3) = \frac{5}{32}$.

### The Complementary Shortcut

Instead of counting the remaining 24 outcomes where $X=-1$, we use the beauty of the complement rule. Since the sum of all probabilities must be 1, we have:

$$P(X=-1) = 1 - [P(X=3) + P(X=4) + P(X=5)]$$

Substituting our values, we get:

$$P(X=-1) = 1 - \frac{5+2+1}{32} = 1 - \frac{8}{32} = \frac{24}{32}$$

### The Expected Value

Finally, we calculate the expected value $E[X] = \sum x_i P(X=x_i)$. Substituting our values:

$$E[X] = (-1)\left(\frac{24}{32}\right) + 3\left(\frac{5}{32}\right) + 4\left(\frac{2}{32}\right) + 5\left(\frac{1}{32}\right)$$

Calculating the numerator: $-24 + 15 + 8 + 5 = 4$.

So, $E[X] = \frac{4}{32} = \frac{1}{8}$.

And there you have it! The **expected value is $\frac{1}{8}$**. It is a small, positive number, suggesting that despite the penalty, the rare occurrences of long streaks of heads pull the average into positive territory.

Animated Solution for Mathematics - Probability: An unbiased coin is tossed 5 times. Suppose that a variable $X$ is assigned the value $k$ when $k$ consecutive heads are obtained for $k = 3, 4, 5$ , otherwise $X$ takes the value $- 1$ . The expected value of $X$ , is

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Visualized Solution

Experiment and Sample Space

Defining the Random Variable $X$

Case $X = 5$

Case $X = 4$

Case $X = 3$

Case $X = - 1$

Expected Value Formula

Substituting the Values

Final Calculation

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The "Exactly" Trap

Case-by-Case Analysis

The Complementary Shortcut

The Expected Value

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