Suppose the probability that a coin toss shows "head" is p, where 0 < p < 1. The coin is tossed repeatedly until the first "head" appears. The expected number of tosses required is
Understanding Coin Toss Expectation
This question asks for the expected number of coin tosses needed to get the first "head", given that the probability of getting a "head" on any single toss is p. This scenario is a classic example of a process modeled by the **Geometric distribution**.
What is the Geometric Distribution?
The Geometric distribution describes the number of Bernoulli trials required to achieve the first success. In this case:
- A single trial is a coin toss.
- Success is defined as getting a "head".
- The probability of success (getting a "head") in a single trial is p.
- The probability of failure (getting a "tail") in a single trial is 1 - p.
- The trials are independent.
Let X be the random variable representing the number of tosses required to get the first "head". The probability mass function for X is given by:
\(P(X=k) = (1-p)^{k-1}p\), for k = 1, 2, 3, ...
This formula means that to get the first success on the k-th trial, you must have k-1 failures followed by one success.
Calculating the Expected Number of Tosses
A key property of the Geometric distribution is its expected value (or mean). The expected value represents the average number of trials needed to achieve the first success over many repetitions of the experiment.
The formula for the expected value, E[X], of a Geometric distribution with probability of success p is:
\(\boldsymbol{E[X] = \frac{1}{p}}\)
Applying the Formula
In this problem, the probability of getting a "head" (success) is given as p. Therefore, the expected number of tosses required to get the first "head" is directly calculated using the formula:
\(\text{Expected tosses} = \frac{1}{p}\)
Analyzing the Options
Let's compare this result with the given options:
- Option 1: \(\frac{1}{p^2}\) - This is the formula for the variance of a Geometric distribution, not the expected value.
- Option 2: \(\frac{{1 - p}}{p}\) - This represents the expected number of *failures* before the first success occurs. If Y = X - 1 is the number of failures, then \(E[Y] = \frac{1-p}{p}\).
- Option 3: \(\frac{{1}}{p}\) - This matches the calculated expected value for the number of tosses until the first success.
- Option 4: \(\frac{{p}}{1 - p}\) - This is not a standard formula associated with the Geometric distribution's expected value or variance.
Therefore, the correct option is the one that provides \(\frac{1}{p}\).
Important Questions from Random Variables Basics
The length of time X, needed by an examinee of competition to complete a 1-hour exam, is a random variable with
PDF \(f(x)=\dfrac{6}{5}(x^2+x);0 \le x \le 1.\) , The value of F(0.5) is:If X follows a binomial distribution with n = 6 and \(p=\dfrac{1}{4}\) then the skewness of X is:
If the customers arrive in a shop in Poisson fashion with parameter λ, the fourth raw moment \(\mu_4^{'}\) for the inter-arrival time is:
A discrete random variable X has the probability functions as:
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f(x)
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2k
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5k
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The value of E(X) is:What percentage of scores falls within three standard deviations from the mean for the normal variate?
