A biology lab stores petri dishes with bacterial cultures in three refrigerators, labeled A, B, and C. Refrigerator A holds 12 dishes: 7 contain strain X and 5 contain strain Y. Refrigerator B holds 18 dishes: 6 contain strain X and 12 contain strain Y. * Refrigerator C holds 10 dishes: 4 contain strain X and the rest contain strain Y. A lab technician will first select a refrigerator at random and then randomly select one dish from that refrigerator. If the selected dish contains strain X, what is the probability that it came from refrigerator A?

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SAT Probability Question 45 | Free SAT Prep | Aniko

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Question 45·Hard·Probability and Conditional Probability

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A biology lab stores petri dishes with bacterial cultures in three refrigerators, labeled A, B, and C.

Refrigerator A holds 12 dishes: 7 contain strain X and 5 contain strain Y.
Refrigerator B holds 18 dishes: 6 contain strain X and 12 contain strain Y.
Refrigerator C holds 10 dishes: 4 contain strain X and the rest contain strain Y.

A lab technician will first select a refrigerator at random and then randomly select one dish from that refrigerator. If the selected dish contains strain X, what is the probability that it came from refrigerator A?

For multi-step probability problems like this, recognize when the question is conditional: phrases like "if it contains X" or "given that" usually mean you want $P(\text{source} \mid \text{outcome})$ . Build a quick mental or written tree: first branch for which refrigerator is chosen, second branch for which strain is selected. Multiply along each branch to get joint probabilities like $P(A \text{ and } X)$ , then add across branches to get the total probability of the outcome (here, getting X). Finally, use the conditional probability formula $P(\text{source}\mid\text{outcome}) = P(\text{source and outcome})/P(\text{outcome})$ , and simplify carefully, watching for arithmetic errors.

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Translate the question into a conditional probability

The phrase "If the selected dish contains strain X, what is the probability that it came from refrigerator A?" is asking for a conditional probability. Think of this as $P(\text{A} \mid X)$ : the probability of A given X has occurred.

Break the process into two stages

First a refrigerator is chosen at random, then a dish is chosen from that refrigerator. For each refrigerator, multiply the probability of choosing that refrigerator by the probability of getting strain X from it to find $P(A \text{ and } X)$ , $P(B \text{ and } X)$ , and $P(C \text{ and } X)$ .

Find the overall chance of getting strain X

Add the three joint probabilities you found to get $P(X)$ , the total probability that the selected dish has strain X, no matter which refrigerator it came from.

Use the conditional probability formula

Use $P(A\mid X) = \dfrac{P(A \text{ and } X)}{P(X)}$ , plugging in your values for $P(A \text{ and } X)$ and $P(X)$ , then simplify the resulting fraction.

Desmos Guide

Compute the probability of getting X from each refrigerator

In Desmos, enter three expressions on separate lines: a = (1/3)*(7/12), b = (1/3)*(6/18), and c = (1/3)*(4/10). These represent $P(A \text{ and } X)$ , $P(B \text{ and } X)$ , and $P(C \text{ and } X)$ respectively.

Find the total probability of X

On a new line, define d = a + b + c. This value is $P(X)$ , the total probability that the selected dish has strain X. If Desmos shows a decimal, tap it to convert it to a fraction.

Form the conditional probability

Now compute a/d. This is $P(A\mid X)$ , the probability that the dish came from refrigerator A given that it contains strain X. Compare the resulting fraction to the answer choices to select the matching one.

Step-by-step Explanation

Find the probability of getting strain X from each refrigerator

The technician chooses each refrigerator with probability $\tfrac{1}{3}$ .

From A: probability of X given A is $7/12$ , so

P(A \text{ and } X) = \frac{1}{3} \cdot \frac{7}{12} = \frac{7}{36}.

From B: probability of X given B is $6/18 = 1/3$ , so

P(B \text{ and } X) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}.

From C: probability of X given C is $4/10 = 2/5$ , so

P(C \text{ and } X) = \frac{1}{3} \cdot \frac{2}{5} = \frac{2}{15}.

Compute the total probability of selecting strain X

Add the probabilities of getting X through each refrigerator:

P(X) = P(A \text{ and } X) + P(B \text{ and } X) + P(C \text{ and } X).

Substitute the values:

P(X) = \frac{7}{36} + \frac{1}{9} + \frac{2}{15}.

Use a common denominator of 180:

$\tfrac{7}{36} = \tfrac{35}{180}$
$\tfrac{1}{9} = \tfrac{20}{180}$
$\tfrac{2}{15} = \tfrac{24}{180}$

P(X) = \frac{35 + 20 + 24}{180} = \frac{79}{180}.

Set up the conditional probability for "from A given X"

We want the probability that the dish came from refrigerator A given that it contains strain X, which is written $P(A\mid X)$ .

The conditional probability formula is

P(A\mid X) = \frac{P(A \text{ and } X)}{P(X)}.

Plug in the values found earlier:

P(A\mid X) = \frac{\tfrac{7}{36}}{\tfrac{79}{180}}.

Simplify the expression to get the final answer

Now simplify

P(A\mid X) = \frac{\tfrac{7}{36}}{\tfrac{79}{180}} = \frac{7}{36} \cdot \frac{180}{79}.

Simplify $\tfrac{180}{36} = 5$ , so

P(A\mid X) = 7 \cdot \frac{5}{79} = \frac{35}{79}.

So the probability that the dish came from refrigerator A, given that it contains strain X, is $\dfrac{35}{79}$ , which corresponds to choice B.

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Step-by-step Explanation

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