Question 7
The records of a shoe manufacturer show that 10% of shoes made are defective.
Assuming independence, the probability of getting 2 defective shoes in a batch of 20 is
Not quite. 0.1937 is the probability of exactly 2 defective shoes in a batch of 10, not 20. The batch size is $n=20$; the 10 belongs to the 10% rate, $p=0.1$.
That’s it. Step through the hints below to see why.
Not quite. 0.3917 is $P(X\le1)$, the chance of at most one defective shoe: $P(X=0)+P(X=1)$. The question asks for exactly 2.
Not quite. 0.6083 is $P(X\ge2)=1-P(X\le1)$, the chance of two or more defective shoes. The question asks for exactly 2, which is a single bar.
This is the question: exactly 2 defective shoes in a batch of 20. That is a single bar, 0.2852, option (B).
This is a batch of 10, not 20. Exactly 2 defective shoes here has probability 0.1937, option (A). Slide the batch size up to 20.
A batch of 19 also gives 0.2852 for exactly 2. That’s a coincidence: $\binom{19}{2}(0.9)^{17}$ and $\binom{20}{2}(0.9)^{18}$ are equal, because $190\times0.9=171$. The question’s batch is 20.
This is at most one defective shoe, the bars for 0 and 1: 0.3917, option (C). It doesn’t include the bar for 2.
This is two or more defective shoes: every bar from 2 up, 0.6083, option (D). Exactly 2 is only the first of those bars.
The orange bars add to . Set the batch size to 20 and pick exactly 2 to match the question.
Each shoe is either defective or not, with the same probability 0.1, independently. There are 20 shoes. Let $X$ be the number of defective shoes in the batch.
$$X\sim\text{Bin}(20,\,0.1)$$
“2 defective shoes” means exactly 2. That is one bar, not a running total like “at most” or “at least”.
$\displaystyle P(X=2)$$\displaystyle =\binom{20}{2}(0.1)^2(0.9)^{18}$
$\displaystyle P(X=2)$$\displaystyle =190\times0.01\times0.9^{18}$$\displaystyle \approx0.2852$
On your calculator, use the binomial probability (pdf) function with $n=20$, $p=0.1$ and $x=2$. The cumulative (cdf) function gives running totals, which is how 0.3917 and 0.6083 turn up.
$$X\sim\text{Bin}(20,\,0.1)$$
$$P(X=2)=\binom{20}{2}(0.1)^2(0.9)^{18}$$
$\displaystyle =190\times0.01\times0.15009$$\displaystyle \approx0.2852$
The probability of exactly 2 defective shoes in a batch of 20 is about 0.2852.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it