To find the probability of Bathsheba hanging the blue painting first and then the green painting, we can break the problem down into two parts.
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Probability of hanging the blue painting first: There are 5 paintings, so the probability of selecting the blue painting first is \( \frac{1}{5} \).
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Probability of hanging the green painting second, given that the blue painting is already hung: After hanging the blue painting, there are 4 paintings left (red, yellow, green, purple). The probability of selecting the green painting next is \( \frac{1}{4} \).
Now, we multiply these two probabilities together to get the overall probability of both events happening:
\[ P(\text{blue first and green second}) = P(\text{blue first}) \times P(\text{green second | blue first}) = \frac{1}{5} \times \frac{1}{4} = \frac{1}{20} \]
Given the options provided, none of them seem to directly correspond to \(\frac{1}{20}\). However, if we consider that the responses may have errors or are formatted differently, the essential calculation remains the same.
In the context of the response options:
- The correct numerical representation as a product of probabilities (like \( \frac{1}{5} \cdot \frac{1}{4} \) yielding \( \frac{1}{20} \)) matches the format better but is not directly visible as you listed it.
Therefore, the final probability of hanging the blue painting first and then the green painting is:
\[ \frac{1}{20} \] (not present in your options).
However, if I had to choose one from the given choices that closely resembles the multiplication of probabilities, it would be the first option, which seems to imply some multiplicative calculation, but constrained by the available selections.