## Related questions with answers

A researcher receives 100 containers of oxygen. Of those containers, 20 have oxygen that is not ionized, and the rest are ionized. Two samples are randomly selected, without replacement, from the lot. a. What is the probability that the first one selected is not ionized? b. What is the probability that the second one selected is not ionized given that the first one was ionized? c. What is the probability that both are ionized? d. How does the answer in part (b) change if samples selected were replaced prior to the next selection?

Solution

Verified$\mathbb{P}(\text{first is not ionized}) = \dfrac{\text{number of not ionized}}{\text{total number}} = \dfrac{20}{100} = \boxed{0.2}.$

$\mathbb{P}\left(\text{second is not ionized} \ \Big \rvert \ \text{first is ionized} \right ) = \dfrac{\text{number of not ionized}}{\text{total number}} = \dfrac{20}{99} \approx \boxed{0.202}.$

$\mathbb{P}(\text{both are ionized}) = \underbrace{\mathbb{P}\left(\text{second is ionized} \ \Big \rvert \ \text{first is ionized} \right)}_{= 79/99} \underbrace{\mathbb{P}(\text{first is ionized})}_{= 80/100} = \dfrac{6320}{9900} \approx \boxed{0.638}.$

If sampling is done with replacement, then we indeed have to adjust solution for part b). Events $\{\text{second is not ionized} \}$ and $\{\text{first is ionized}\}$ become independent, therefore we have:

$\mathbb{P}\left(\text{second is not ionized} \ \Big \rvert \ \text{first is ionized} \right ) = \mathbb{P}(\text{second is not ionized}) = \dfrac{\text{number of not ionized}}{\text{total number}} = \dfrac{20}{100} = \boxed{0.20}.$

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