Part 4 or 6 In a certain game of chance, a wheel consists of 48 slots numbered \( 00,0,1,2, \ldots, 46 \). To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. Complete parts (a) through (c) below. (a) Determine the sample space. Choose the correct answer below. A. The sample space is \( \{00,0,1,2, \ldots, 46\} \). B. The sample space is \( \{1,2, \ldots, 46\} \). C. The sample space is \( (00) \). D. The sample space is \( \{00,0\} \). (b) Determine the probability that the metal ball falls into the slof marked 4. Interpret this probability. The probability that the metal ball falls into the slot marked 4 is \( \square \) 0.0208 . (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Type a whole number.) A. If the wheet is spun 1000 times, it is expected that about 21 of those times result in the ball landing in slot 4 . B. If the wheel is spun 1000 times, it is expected that exactly \( \square \) of those times result in the ball not landing in slot 4 . (c) Determine the probability that the metal ball lands in an odd slot. Interpret this probability. The probabillty that the metal ball lands in an odd slot is \( \square \) . (Round to four decimal places as needed.)
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The sample space includes all possible outcomes of the game, which consists of 48 slots ranging from \( 00 \) to \( 46 \). Therefore, the correct answer is **A**: The sample space is \( \{00, 0, 1, 2, \ldots, 46\} \). For part (a), the probability that the metal ball falls into the slot marked 4 is calculated as the number of favorable outcomes (which is 1, landing in slot 4) divided by the total outcomes (which is 48). Thus, the probability is \( \frac{1}{48} \approx 0.0208 \). For interpreting this probability, the correct choice is **A**: If the wheel is spun 1000 times, it is expected that about 21 of those times result in the ball landing in slot 4 (since \( 1000 \times 0.0208 \approx 20.833 \), which rounds to about 21). For part (c), the odd slots in the sample space are \( 1, 3, 5, 7, \ldots, 45 \), totaling 24 odd numbers (since the sequence of odd numbers from 1 to 46 is an arithmetic series). The probability of landing in an odd slot is therefore \( \frac{24}{48} = 0.5 \). This means that if the metal ball is spun, there is a 50% chance it will land in any odd-numbered slot, making it a fair game where the outcome is equally likely!