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One pip (small dot on the face of a die) is randomly removed from a standard eight-sided die, where each pip has an equal probability of being removed (this means for example, the face with 8 pips has a greater probability of losing a pip compared to the face with 1 pip). What is the probability that the top face will be an even number when rolling this new die?

User Ingham
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Final answer:

The probability of rolling an even number on a modified eight-sided die after removing a pip depends on which face the pip is removed from. Without specific details on which pip is removed, we cannot calculate an exact probability. However, removal from an even face decreases the chance of getting an even number, while removal from an odd face keeps the probability unchanged.

Step-by-step explanation:

The question asks for the probability that the top face of a modified eight-sided die will show an even number after one pip has been removed from a random face. To calculate this, first understand that an eight-sided die (often called an octahedron in dice terminology) has faces numbered from 1 to 8. The even faces are 2, 4, 6, and 8.

Before removing a pip, each face has an equal probability of 1/8. However, the removal of one pip affects the probabilities in a non-uniform way; faces with more pips have a higher chance of having one removed.

Assuming we have removed one pip, there are two scenarios that can occur that will affect the even-numbered faces:

  1. If the removed pip comes from an even-numbered side, then one potential even result is lost and the remaining even sides become more likely.
  2. If the removed pip comes from an odd-numbered side, the probability of rolling an even number remains the same.

The exact probability depends on which pip is removed, but since we are not given specifics, we cannot provide a single numeric answer. Instead, we understand that the removal of a pip from an even face will decrease the probability of rolling an even number, and the removal from an odd face will not change the even probability.

User Stano
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