Question

an urn contains different colored marbles. the probability of drawing two green marbles from the urn without replacement is $\frac{3}{20}$, and the probability of drawing one green marble is $\frac{2}{5}$.
what is the probability of drawing a second green marble, given that the first marble is green?
$\frac{3}{8}$
$\frac{3}{30}$
$\frac{1}{5}$
$\frac{1}{2}$

Explanation:

Step1: Recall conditional probability formula

The formula for conditional probability is \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). Let \(A\) be the event of drawing a green marble first and \(B\) be the event of drawing a green marble second. Then \(P(A\cap B)\) is the probability of drawing two green marbles (without - replacement), and \(P(A)\) is the probability of drawing one green marble.

Step2: Substitute the given values

We are given that \(P(A\cap B)=\frac{3}{20}\) and \(P(A)=\frac{2}{5}\).

$$P(B|A)=\frac{\frac{3}{20}}{\frac{2}{5}}$$

Step3: Simplify the fraction

When dividing by a fraction, we multiply by its reciprocal. So \(\frac{\frac{3}{20}}{\frac{2}{5}}=\frac{3}{20}\times\frac{5}{2}\).

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Answer:

\(\frac{3}{8}\)