Digital data received from a sensor can fill up 0 to 32 buffers. Let the sample space be S = {0, 1, 2, .........., 32} where the sample j denote that j of the buffers are full and \(p\left( i \right) = \frac{1}{{561}}\left( {33 - i} \right)\)
0.515
Understanding the Sensor Data Probability Problem
The problem describes a scenario involving digital data from a sensor filling up buffers. We are given the possible number of full buffers and a rule to calculate the probability for each specific number of full buffers.
- The sample space \(S\) represents all possible outcomes, which is the number of buffers that can be full. According to the problem, this number can range from 0 to 32. So, \(S = \{0, 1, 2, \ldots, 32\}\).
- The total number of possible outcomes in this sample space is \(32 - 0 + 1 = 33\).
- The probability function \(p(j)\) gives the probability that exactly \(j\) buffers are full. The formula is given as \(p(i) = \frac{1}{{561}}\left( {33 - i} \right)\). Using \(j\) for the number of buffers, the probability is \(p(j) = \frac{1}{561}(33 - j)\) for any \(j \in S\).
Defining the Event A: Even Number of Full Buffers
The problem asks for the probability of event A, which is defined as the event that an even number of buffers are full. To find the probability of event A, we first need to identify all the outcomes in the sample space S that are part of event A.
- Event A includes all numbers from the sample space \(S = \{0, 1, 2, \ldots, 32\}\) that are even.
- So, the set of outcomes for event A is \(A = \{0, 2, 4, 6, \ldots, 32\}\).
Calculating the Probability of Event A, p(A)
For a discrete probability distribution, the probability of an event is the sum of the probabilities of all the individual outcomes that make up that event. Therefore, to find the probability of event A, denoted as \(P(A)\) or \(p(A)\), we need to sum the probabilities \(p(j)\) for all \(j\) belonging to set A.
\[ P(A) = \sum_{j \in A} p(j) = p(0) + p(2) + p(4) + \cdots + p(32) \]Substitute the given probability formula \(p(j) = \frac{1}{561}(33 - j)\) for each term:
\[ P(A) = \frac{1}{561}(33 - 0) + \frac{1}{561}(33 - 2) + \frac{1}{561}(33 - 4) + \cdots + \frac{1}{561}(33 - 32) \]We can factor out the common constant \(\frac{1}{561}\) from each term:
\[ P(A) = \frac{1}{561} \left[ (33 - 0) + (33 - 2) + (33 - 4) + \cdots + (33 - 32) \right] \] \[ P(A) = \frac{1}{561} \left[ 33 + 31 + 29 + \cdots + 1 \right] \]The terms inside the square brackets form a sequence where the difference between consecutive terms is constant. This is an arithmetic progression.
- The first term (\(a_1\)) is 33.
- The common difference (\(d\)) is \(31 - 33 = -2\).
- The last term (\(a_n\)) is 1.
To find the number of terms (\(n\)) in this arithmetic progression, we use the formula for the n-th term of an arithmetic progression: \(a_n = a_1 + (n-1)d\).
\[ 1 = 33 + (n - 1)(-2) \] \[ 1 - 33 = -2(n - 1) \] \[ -32 = -2(n - 1) \]Divide both sides by -2:
\[ \frac{-32}{-2} = n - 1 \] \[ 16 = n - 1 \]Solve for n:
\[ n = 16 + 1 \] \[ n = 17 \]There are 17 terms in the sum \(33 + 31 + 29 + \cdots + 1\).
Now, we calculate the sum of this arithmetic progression using the formula for the sum of n terms: \(S_n = \frac{n}{2}(a_1 + a_n)\).
\[ \text{Sum} = \frac{17}{2}(33 + 1) \] \[ \text{Sum} = \frac{17}{2}(34) \] \[ \text{Sum} = 17 \times 17 \] \[ \text{Sum} = 289 \]So, the sum of the probability components for event A is 289.
Finally, we substitute this sum back into the expression for \(P(A)\):
\[ P(A) = \frac{1}{561} \times \text{Sum} \] \[ P(A) = \frac{1}{561} \times 289 \] \[ P(A) = \frac{289}{561} \]To get the decimal value, we perform the division:
\[ P(A) \approx 0.515151... \]Rounding to three decimal places, the probability \(P(A)\) is approximately 0.515.
Revision Table: Understanding Probability Concepts
| Concept | Explanation |
|---|---|
| Sample Space (S) | The set of all possible outcomes in a probability experiment (here, number of full buffers from 0 to 32). |
| Probability Distribution (p(j)) | A rule or function that assigns a probability value to each outcome in the sample space. The sum of all probabilities must be 1. |
| Event (A) | A specific subset of the sample space (here, having an even number of full buffers). |
| Probability of an Event (P(A)) | The sum of the probabilities of all the individual outcomes that make up the event. |
| Arithmetic Progression | A sequence where the difference between consecutive terms is constant. The sum can be calculated using specific formulas. |
Additional Information: Discrete Probability Distributions
This problem deals with a discrete probability distribution because the sample space is finite (countable). A discrete probability distribution assigns a probability to each distinct outcome.
A key property of any probability distribution is that the sum of the probabilities of all possible outcomes must equal 1. For the given probability function \(p(j) = \frac{1}{561}(33 - j)\) over \(S = \{0, 1, \ldots, 32\}\), we can verify this:
\[ \sum_{j=0}^{32} p(j) = \sum_{j=0}^{32} \frac{1}{561}(33 - j) \] \[ = \frac{1}{561} \sum_{j=0}^{32} (33 - j) \]The sum \(\sum_{j=0}^{32} (33 - j)\) is \( (33-0) + (33-1) + \cdots + (33-32) = 33 + 32 + \cdots + 1 \). This is the sum of the first 33 positive integers, which is given by the formula \(\frac{n(n+1)}{2}\) with \(n=33\).
\[ \text{Sum} = \frac{33(33+1)}{2} = \frac{33 \times 34}{2} = 33 \times 17 = 561 \]So, the total probability is \(\frac{1}{561} \times 561 = 1\), confirming that \(p(j)\) is a valid probability function over the sample space S.
Calculating the probability of an event like 'even number of buffers' involves summing probabilities of specific outcomes, as demonstrated in the solution above using properties of arithmetic progressions.
Important Questions from Discrete Distributions
The value of a and b so that the following is probability mass function
X: 0 1 2 P(X = x): 3a 3b 4b with mean 1.1, is:
If X is a Poisson random variate with mean 3, then P(|X- 3| < 1) will be:
Let x ∼ N(μ, σ2) If μ2 = σ2, (μ > 0), then the value of P(X < -μ | X < μ) in terms of cumulative function N (0, 1) is:
Consider a binomial random variable X. If X1, X2,...Xn are independent and identically distributed samples from the distribution of X with sum \(Y = \mathop \sum \limits_{i = 1}^n {X_i}\) then the distribution of Y as n → ∞ can be approximated as.
- Identify the generic probability density function that corresponds with discrete random variables.
