Probability Calculation for Defective Balls in a Sample

The probability that a ball drawn at random from an urn of balls is defective is 0.3. If a sample of 5 balls is taken, what is the probability that it will contain no defective balls or four or five defective balls? This article will explore the solution step-by-step using the binomial probability formula.

Understanding the Problem

Given that the probability of drawing a defective ball is 0.3, we can denote this probability as (p 0.3). Therefore, the probability of drawing a non-defective ball is (1 - p 0.7). The total number of balls in the sample is 5, which we denote as (n 5).

Binomial Probability Formula

The binomial probability formula is given by:

$$ P(X k) binom{n}{k} p^k (1-p)^{n-k} $$

Where:

$$n$$ is the number of trials (the number of balls drawn) in this case, (n 5). $$k$$ is the number of successful trials (the number of defective balls) in this case, (k) can be 0, 4, or 5. $$p$$ is the probability of success on an individual trial, in this case, (p 0.3). $$1 - p$$ is the probability of failure on an individual trial, in this case, (1 - p 0.7). $$binom{n}{k}$$ is the binomial coefficient, which gives the number of ways to choose (k) successes in (n) trials.

The binomial coefficient (binom{n}{k}) is calculated as follows:

$$ binom{n}{k} frac{n!}{k!(n-k)!} $$

Case I: No Defective Balls

Let's compute the probability that there are no defective balls in the sample (i.e., (k 0)).

Using the binomial probability formula:

$$ P(X 0) binom{5}{0} 0.3^0 (1-0.3)^{5-0} $$

Calculating each component:

$$binom{5}{0} 1$$ $$0.3^0 1$$ $$(1-0.3)^5 0.7^5 0.16807$$

Therefore, the probability (P(X 0)) is:

$$ P(X 0) 1 cdot 1 cdot 0.16807 0.16807 $$

Case II: Four or Five Defective Balls

To find the probability that there are four or five defective balls in the sample (i.e., (k 4) or (k 5)), we need to compute the individual probabilities for (k 4) and (k 5) and then sum them up.

Sub-case: Four Defective Balls (k 4)

Using the binomial probability formula:

$$ P(X 4) binom{5}{4} 0.3^4 (1-0.3)^{5-4} $$

Calculating each component:

$$binom{5}{4} 5$$ $$0.3^4 0.0081$$ $$(1-0.3)^1 0.7$$

Therefore, the probability (P(X 4)) is:

$$ P(X 4) 5 cdot 0.0081 cdot 0.7 0.02835 $$

Sub-case: Five Defective Balls (k 5)

Using the binomial probability formula:

$$ P(X 5) binom{5}{5} 0.3^5 (1-0.3)^{5-5} $$

Calculating each component:

$$binom{5}{5} 1$$ $$0.3^5 0.00243$$ $$(1-0.3)^0 1$$

Therefore, the probability (P(X 5)) is:

$$ P(X 5) 1 cdot 0.00243 cdot 1 0.00243 $$

Combining the probabilities for (k 4) and (k 5), we get:

$$ P(X 4 text{ or } X 5) P(X 4) P(X 5) 0.02835 0.00243 0.03078 $$

Summary of Results

The probability of no defective balls: (0.16807) The probability of four or five defective balls: (0.03078)

This concludes our detailed step-by-step solution using the binomial probability formula to find the required probabilities.