probability of random selection

These “there are 6 red gumballs and 4 blue in a jar, you pick two, what’s the likelihood of getting two blue ones?” type questions are a type of conditional probability.

In the above example, you’d have a 4/10 chance the first time, and then a 3/9 chance the second. So $\frac{4}{10}\cdot \frac{3}{9}=.1\bar{3}$.

Solve this problem

• DONE Completed: 2024-09-11 Thanks chatgpt A pharmaceutical company receives large shipments of ibuprofen tablets and uses this acceptance sampling plan: randomly select and test 30 tablets, then accept the whole batch if there is at most one that doesn’t meet the required specifications. If a particular shipment of thousands of ibuprofen tablets actually has a 9% rate of defects, what is the probability that this whole shipment will be accepted?

P(accept shipment) = .9390 Actual: .2343

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This uses binomial distribution.

Given we have 30 tablets, n=30. We choose k=1. Given a probability of defect of 0.09, success likelihood is 1-.09=0.91

So probability of 1 positive among the 30 options is:

$\begin{array}{rl}& =\begin{array}{c}n\\ k\end{array}\cdot p^k\cdot (1-p{)}^{n-k}\\ & =\begin{array}{c}30\\ 1\end{array}\cdot .09^1\cdot (1-.09{)}^{30-1}\\ & =30\cdot .09^1\cdot (.91{)}^{29}\\ & =30\cdot .09\cdot .065\\ & =30\cdot .09\cdot .065\\ & =.1755\end{array}$

And the probability there are no errors:

$\begin{array}{rl}& =\begin{array}{c}n\\ k\end{array}\cdot p^k\cdot (1-p{)}^{n-k}\\ & =\begin{array}{c}30\\ 0\end{array}\cdot .09^0\cdot (1-.09{)}^{30-0}\\ & =1\cdot 1\cdot (.91{)}^{30}\\ & =1\cdot 1\cdot .059\\ & =.059\end{array}$

so in total: .1755 + .059 = .2345