Question

Let the random variable \(\mathrm{X}\) represent the number of defective radios in a shipment of four radios to a local appliance store. Assume that each radio is equally likely to be defective or non-defective, hence the probability that a radio is defective is \(\mathrm{p}=1 / 2\). Also assume whether or not each radio is defective or non-defective is indipendent of the status of the other radios. Find the expected number of defective radios.

Short answer

The expected number of defective radios in a shipment of four, given that each radio is equally likely to be defective or non-defective, is 2 radios. This is found by calculating the expected value of the binomially distributed random variable \(X\), which represents the number of defective radios in the shipment. The expected value \(E(X)\) is calculated as the sum of the product of each value of \(X\) and its corresponding probability. In this case, \(E(X) = 0*P(X=0) + 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + 4*P(X=4) = 2\).

Step-by-step solution

Determine the probability distribution of X

Since the number of trials (n) is 4, the probability of success (p) is 1/2, and the status of each radio is independent, we can use the binomial distribution to find the probabilities of each value of X: \[P(X = k) = C(n, k) p^k (1 - p)^{n-k}\] Where C(n, k) is the binomial coefficient, which represents the number of ways to choose k successes from n trials.

Calculate the probabilities for X=0,1,2,3,4

We will apply the binomial formula to find the probabilities of each value of X: \(P(X=0) = C(4,0) (1/2)^0 (1/2)^4= 1*(1)*(1/16) = 1/16\) \(P(X=1) = C(4,1) (1/2)^1 (1/2)^3= 4*(1/2)*(1/8) = 1/4\) \(P(X=2) = C(4,2) (1/2)^2 (1/2)^2= 6*(1/4)*(1/4) = 3/8\) \(P(X=3) = C(4,3) (1/2)^3 (1/2)^1= 4*(1/8)*(1/2) = 1/4\) \(P(X=4) = C(4,4) (1/2)^4 (1/2)^0= 1*(1/16)*(1) = 1/16\)

Calculate the expected value of X:

Finally, we will calculate the expected value (mean) of X: \[E(X) = \sum_{k=0}^{4} k P(X=k)\] \[E(X)= 0*P(X=0) + 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + 4*P(X=4)\] Plugging in the probabilities we calculated earlier: \[E(X) = 0*(1/16) + 1*(1/4) + 2*(3/8) + 3*(1/4) + 4*(1/16)\] \[E(X) = 0 + 1/4 + 3/4 + 3/4 + 1/4\] \[ E(X) = 2\] The expected number of defective radios in the shipment is 2.

Key concepts

Expected Value

The concept of expected value is crucial when dealing with random variables and probability distributions. It's essentially the weighted average or mean of all possible values that a random variable can take. In probability, the expected value gives us a measure of the center of a distribution, reflecting what you might expect to happen on average when repeating the experiment many times.

For a binomial distribution, calculating the expected value is straightforward. The formula for the expected value, denoted as \(E(X)\), is given by:

• \(E(X) = \sum_{k=0}^{n} k \cdot P(X=k)\)

where \(k\) is each possible outcome, and \(P(X=k)\) is the probability of \(X\) being equal to \(k\).

In our exercise, \(X\) represents the number of defective radios in a shipment of four radios. Here, we calculated the expected value of \(X\) by considering the product of each possible number of defective radios and their corresponding probabilities. After working through each term, we find that the expected number of defective radios in this case is 2. This tells us that if we carried out this scenario repeatedly, on average, we would expect each shipment to contain two defective radios.

Probability Distribution

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes for an experiment. In the context of a binomial distribution, we are specifically interested in the probability distribution of a binomial random variable.

For the binomial distribution, the probability of exactly \(k\) successes in \(n\) independent Bernoulli trials, each with a probability \(p\) of success, is given by the formula:

• \(P(X = k) = C(n, k) p^k (1 - p)^{n-k}\)

Here, \(C(n, k)\) is the binomial coefficient, \(p\) is the probability of success, \(1 - p\) is the probability of failure, and \(n\) is the number of trials.

For our random variable \(X\), representing defective radios, the probability distribution assigns probabilities to each possible number of defective radios: 0, 1, 2, 3, or 4. With a given probability of failure or defect \(p = 1/2\), we've shown how to compute these probabilities and establish the complete probability distribution for the scenario.

Binomial Coefficient

The binomial coefficient is a fundamental component in the calculation of probabilities within a binomial distribution. It determines the number of ways to choose \(k\) successes from \(n\) trials and is often denoted as \(C(n, k)\) or \(\binom{n}{k}\).

Mathematically, the binomial coefficient is expressed as:

• \(C(n, k) = \frac{n!}{k!(n-k)!}\)

where \(!\) (factorial) means to multiply all whole numbers from "n" down to 1.

In the context of our radio shipment problem, the binomial coefficient helps us know how likely it is to have a certain number of defective radios. For example, \(C(4, 2) = 6\) signifies that there are 6 ways to have exactly 2 defective radios out of 4. This is a key step in determining the probability of each scenario within the binomial distribution.