SAT Probability Question 27 | Free SAT Prep | Aniko

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Question 27·Hard·Probability and Conditional Probability

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A company uses two machines, Machine A and Machine B, to manufacture identical metal pins. Machine A produces 40% of the pins, and Machine B produces the rest. Historical data show that 3% of the pins produced by Machine A are defective, whereas 5% of the pins produced by Machine B are defective.

A pin is selected at random from one day’s production and is found to be defective. What is the probability that the defective pin was produced by Machine A?

For problems asking for the probability that an outcome came from a particular group given some condition (like being defective), set up a clear structure: write down each group’s share of the total and its conditional probability for the condition, multiply to get the joint probabilities for each group with the condition, sum these to get the total probability of the condition, and finally use $P(\text{group}\mid\text{condition}) = \dfrac{P(\text{group and condition})}{P(\text{condition})}$ . To work quickly under time pressure, you can also imagine a convenient total number of items (like 1,000 pins), compute how many defective pins each machine would produce, and take the fraction from the machine of interest over the total defective pins.

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Identify the given probabilities

Write down the probability that a pin comes from each machine, and the probability that a pin from each machine is defective. Label them as $P(A)$ , $P(B)$ , $P(D\mid A)$ , and $P(D\mid B)$ .

Find the overall chance that a pin is defective

First compute the probability that a pin is defective and from Machine A, and the probability that it is defective and from Machine B. Then add these to get the total probability that a pin is defective, $P(D)$ .

Use conditional probability the right way around

You want $P(\text{Machine A} \mid \text{defective})$ . Use the formula $P(A\mid D) = \dfrac{P(A\cap D)}{P(D)}$ , and be careful not to mix it up with $P(D\mid A)$ , which is given in the problem.

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Compute the conditional probability numerically

In Desmos, type the expression (0.4*0.03)/(0.4*0.03 + 0.6*0.05) into an empty line. Look at the numerical output that appears to the right.

View the exact fraction form

Click or tap on the decimal result and, if Desmos offers a fraction form, note that simplified fraction. Then compare that fraction to the answer choices to see which option matches the value you computed.

Step-by-step Explanation

Translate the wording into probabilities

Let event $A$ be "pin was made by Machine A" and $B$ be "pin was made by Machine B". Let $D$ be "pin is defective".

From the problem:

$P(A) = 40\% = 0.40$
$P(B) = 60\% = 0.60$ (the rest)
$P(D\mid A) = 3\% = 0.03$ (defective given it’s from A)
$P(D\mid B) = 5\% = 0.05$ (defective given it’s from B)

We are asked to find $P(A\mid D)$ : the probability a pin came from A given that it is defective.

Find how likely a defective pin is from each machine

First find the probability that a randomly chosen pin is both from a machine and defective.

From Machine A and defective:

P(A \cap D) = P(A)\cdot P(D\mid A) = 0.40 \times 0.03 = 0.012

From Machine B and defective:

P(B \cap D) = P(B)\cdot P(D\mid B) = 0.60 \times 0.05 = 0.030

Now add these to get the overall probability that a pin is defective (from either machine):

P(D) = P(A \cap D) + P(B \cap D) = 0.012 + 0.030 = 0.042

Apply the conditional probability formula

The formula for conditional probability is

P(A\mid D) = \frac{P(A \cap D)}{P(D)}.

We already found:

$P(A \cap D) = 0.012$
$P(D) = 0.042$

Substitute these values:

P(A\mid D) = \frac{0.012}{0.042}.

Simplify the fraction and match the answer choice

Simplify $\dfrac{0.012}{0.042}$ by writing it as a fraction with whole numbers:

\frac{0.012}{0.042} = \frac{12}{42}.

Now reduce $\frac{12}{42}$ by dividing numerator and denominator by their greatest common divisor, $6$ :

\frac{12}{42} = \frac{12\div 6}{42\div 6} = \frac{2}{7}.

So the probability that a randomly chosen defective pin was produced by Machine A is $\dfrac{2}{7}$ , which corresponds to choice D.

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Step-by-step Explanation

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Step-by-step Explanation