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\( 1< \) A date in the month of April is chosen at random. Find the probability of the event of the date's being each of the following. \( \begin{array}{ll}\text { a. April } 3 & \text { b. April } 31 \\ \begin{array}{l}\text { a. The probability of the event of the date's being April } 3 \text { is } \frac{1}{30}\end{array} \\ \text { (Type an integer or a simplified fraction.) } \\ \text { b. The probability of the event of the date's being April } 31 \text { is } \\ \text { (Type an integer or a simplified fraction.) }\end{array} \)

Ask by West Mitchell. in the United States
Feb 26,2025

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Answer

a. The probability of the date being April 3 is 1/30. b. The probability of the date being April 31 is 0.

Solution

April has 30 days. a. Since April 3 is one specific day out of 30,   Probability = 1/30. b. April 31 does not exist, so the probability is 0.

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Since April has 30 days, let's take a look at the probabilities: a. The probability of the event of the date's being April 3 is \( \frac{1}{30} \), as there is 1 favorable outcome (April 3) out of 30 possible dates. b. The probability of the event of the date's being April 31 is \( 0 \). This is because April has only 30 days, making it impossible for the 31st of April to occur—no favorable outcomes here!

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