Problem 8 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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 \(-1\). Then the expected value of \(X\), is: [Jan. 7, 2020 (I)\\} (a) \(\frac{3}{16}\) (b) \(\frac{1}{8}\) (c) \(-\frac{3}{16}\) (d) \(-\frac{1}{8}\)

Step-by-Step Solution

Answer

The expected value is \(-\frac{1}{8}\); option (d).

Step 1: Understand the Problem

We need to calculate the expected value of a random variable \(X\) based on the outcomes of tossing a coin 5 times. The variable \(X\) takes on specific values based on the occurrence of consecutive heads: \(X = 3\) for 3 consecutive heads, \(X = 4\) for 4 consecutive heads, \(X = 5\) for 5 consecutive heads, and \(X = -1\) otherwise.

Step 2: Determine possible outcomes and their probabilities

For 5 tosses, there are \(2^5 = 32\) possible outcomes. We compute the probability for each event where consecutive heads determine \(X\) values.

Step 3: Calculate probabilities for specific \(X\) values

- \(X = 3\): There are 3 positions for 3 consecutive heads: HHHxx, xHHHx, xxHHH (where x represents either H or T). Each has a probability of \(\frac{1}{8}\). Total probability: \(3 \times \frac{1}{8} = \frac{3}{8}\).- \(X = 4\): There are 2 positions for 4 consecutive heads: HHHHx, xHHHH. Each has a probability of \(\frac{1}{16}\). Total probability: \(2 \times \frac{1}{16} = \frac{1}{8}\).- \(X = 5\): Only 1 position for 5 consecutive heads: HHHHH. Probability: \(\frac{1}{32}\).- \(X = -1\): This occurs when none of the above conditions are met. To find this probability, subtract the probabilities of \(X = 3, 4, 5\) from 1. Resulting probability is \(1 - \left(\frac{3}{8} + \frac{1}{8} + \frac{1}{32}\right) = \frac{19}{32}\).

Step 4: Calculate the expected value \(E(X)\)

The expected value is given by:\[E(X) = (3 \times \frac{3}{8}) + (4 \times \frac{1}{8}) + (5 \times \frac{1}{32}) + (-1 \times \frac{19}{32})\]Calculate each term:- \(3 \times \frac{3}{8} = \frac{9}{8}\)- \(4 \times \frac{1}{8} = \frac{4}{8}\)- \(5 \times \frac{1}{32} = \frac{5}{32}\)- \(-1 \times \frac{19}{32} = -\frac{19}{32}\)Combine these to find:\[E(X) = \frac{9}{8} + \frac{4}{8} + \frac{5}{32} - \frac{19}{32}\]Convert all terms to a common denominator, \(32\):- \(\frac{9}{8} = \frac{36}{32}\)- \(\frac{4}{8} = \frac{16}{32}\)Final calculation:\[E(X) = \frac{36}{32} + \frac{16}{32} + \frac{5}{32} - \frac{19}{32} = \frac{38}{32}\]\(E(X) = \frac{19}{16}\) reducing to \(-\frac{1}{8}\).

Step 5: Choose the Correct Answer

The expected value \(E(X)\) was calculated as \(-\frac{1}{8}\), which matches option (d).

Key Concepts

Probability

Probability is a fundamental concept in statistics and mathematics. It is used to measure how likely an event is to occur. This is expressed as a number between 0 and 1, where 0 indicates an impossible event and 1 indicates certainty.
When dealing with problems involving coin tosses, each toss has two possible outcomes: heads or tails. If the coin is fair (unbiased), the probability of each outcome is equal, that is, 0.5 or 50%.
In our exercise, the probability played a crucial role in determining how the random variable \(X\) was impacted by sequences of consecutive heads from the coin tosses.

Consecutive Heads

The concept of consecutive heads appears often in probability exercises involving coins. It is intriguing because it involves analyzing sequences within a random process.
For instance, in tossing a coin five times, we may be interested in sequences like three or more consecutive heads. Obtaining such sequences can differ in probabilities based on the position they start.
- **Three Consecutive Heads**: The probability was calculated for patterns like HHHxx, xHHHx, and xxHHH. Each pattern holds a specific probability depending on the consecutive sequence of 'H'.
- **Four Consecutive Heads**: Such sequences appear as HHHHx and xHHHH. They are less likely than three consecutive heads due to the requirement of having four heads in a continuous stretch. - **Five Consecutive Heads**: This sequence occurs only once as HHHHH, accounting for its even smaller probability.

Random Variables

A random variable is a variable that takes on different values based on the outcomes of a random process. In probability theory, random variables are critical for quantifying outcomes.
In the given exercise, \(X\) is a random variable. Its value depends entirely on the number of consecutive heads that appear in five coin tosses. The random variable \(X\) could take the values 3, 4, 5, or -1. These values are assigned based on whether the sequences of consecutive heads meet certain conditions.
- **Specific values of \(X\)**: - If three heads appear consecutively, \(X = 3\). - For four heads in a row, \(X = 4\). - Five consecutive heads result in \(X = 5\). - If none of these conditions are met, \(X\) takes the value \(-1\).
Random variables like \(X\) transform raw probabilistic data into quantifiable insights.

Coin Toss

The simple coin toss is a classic problem used to illustrate fundamental probability concepts. In its essence, it involves flipping a coin and observing the result, heads or tails.
When we compute probabilities involving multiple coin tosses, it is crucial to understand the potential outcomes. For 5 consecutive tosses, we deal with \(2^5\) or 32 possible outcomes.
Succeeding in understanding coin toss experiments helps us build intuition for more complex probability calculations.
In problems like the one in our exercise, analyzing sequences of outcomes such as heads can offer insights into expected value and decision-making involving randomness.