3 A special deck of cards has 5 green cards, and 3 yellow cards. The green cards are numbered 1, 2, 3, 4, and 5. The yellow cards are numbered 1, 2, and 3. The cards are well shuffled and you randomly draw one card. \( G= \) card drawn is green \( \mathrm{E}= \) card drawn is even-numbered a. How many elements are there in the sample space? \( \square \) b. \( P(E)= \) \( \square \) (Round to. 4 decimal places) Hint: Sample Space and Probability Video on Probability [] [+] Textbook Question Help: Message instructor Submit All Parts

**a. Finding the number of elements in the sample space** - There are 5 green cards and 3 yellow cards. - Therefore, the total number of cards is \[ 5 + 3 = 8. \] Thus, the number of elements in the sample space is \( 8 \). --- **b. Calculating \( P(E) \) where \( E \) is the event that the drawn card is even-numbered** 1. **Determine the even-numbered cards:** - For green cards (numbered 1 to 5), the even numbers are 2 and 4. This gives us 2 even-numbered green cards. - For yellow cards (numbered 1 to 3), the even number is 2. This gives us 1 even-numbered yellow card. 2. **Total number of even-numbered cards:** \[ 2 + 1 = 3. \] 3. **Calculate the probability \( P(E) \):** \[ P(E) = \frac{\text{Number of even-numbered cards}}{\text{Total number of cards}} = \frac{3}{8}. \] 4. **Convert the probability to decimal form (rounded to 4 decimal places):** \[ \frac{3}{8} = 0.3750. \] Thus, \( P(E) = 0.3750 \).