Question

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question:
in a shipment of toys from a manufacturer, the probability that a toy is defective is \\(\frac{1}{50}\\). if marie selects 2 toys from a shipment, what is the probability that both toys are defective?
\\(\frac{1}{25}\\)
\\(\frac{1}{50}\\)
\\(\frac{1}{100}\\)
\\(\frac{1}{2,500}\\)

Explanation:

Step1: Identify Probability Rule

For independent events (assuming selecting one toy doesn't affect the other's defectiveness), the probability of both events occurring is the product of their individual probabilities. Here, probability of one defective toy is $\frac{1}{50}$, so for two, we multiply $\frac{1}{50} \times \frac{1}{50}$.

Step2: Calculate the Product

$\frac{1}{50} \times \frac{1}{50} = \frac{1 \times 1}{50 \times 50} = \frac{1}{2500}$. Wait, but the options have $\frac{1}{25}$, $\frac{1}{50}$, $\frac{1}{100}$, $\frac{1}{2500}$. Wait, maybe I misread the probability. Wait, the problem says "the probability that a toy is defective is $\frac{1}{50}$"? Wait, no, maybe it's $\frac{1}{5}$? Wait, no, the user's image: let's recheck. Wait, the probability is $\frac{1}{50}$? Wait, no, maybe it's a typo, but the calculation: if probability of defective is $p$, then probability both defective is $p \times p$. So if $p = \frac{1}{50}$, then $p^2 = \frac{1}{2500}$, which is one of the options (the last one: $\frac{1}{2500}$). Wait, but let's confirm. The options are $\frac{1}{25}$, $\frac{1}{50}$, $\frac{1}{100}$, $\frac{1}{2500}$. So the correct calculation is multiplying the probability of each event (since selecting two toys, assuming independence, the probability both are defective is the product of each being defective). So $\frac{1}{50} \times \frac{1}{50} = \frac{1}{2500}$.

Answer:

$\frac{1}{2500}$ (the last option, e.g., if options are labeled as A. $\frac{1}{25}$, B. $\frac{1}{50}$, C. $\frac{1}{100}$, D. $\frac{1}{2500}$, then D. $\frac{1}{2500}$)