Possible Winnings from the Ring Toss at a Carnival

Let's define the event carefully: - There are **6 bottles**, each throw, you have a chance to ring any one of them. - **You have 3 tries**. - You win the teddy bear **if you ring any one bottle at least once in 3 tries** (since the game says "on a bottle in three tries"). Let’s assume **each throw is independent**, and **each bottle is a valid target**. You only need to succeed once in the 3 tries. Unless stated otherwise, let’s ALSO assume for this standard setup: - You can pick **which bottle to aim at** each time. - Each throw has an equal chance of success (random toss). The **probability of hitting a bottle in one throw** is: \[ P(\text{hit on one throw}) = \frac{1}{6} \] Therefore, \[ P(\text{miss on one throw}) = 1 - \frac{1}{6} = \frac{5}{6} \] **Probability of missing all three tries:** \[ P(\text{miss all 3}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \] **Probability of winning (getting at least one ring in):** \[ P(\text{win}) = 1 - P(\text{miss all 3}) = 1 - \frac{125}{216} = \frac{91}{216} \] To convert to a percentage: \[ \frac{91}{216} \approx 0.4213 \] So, as a percentage: \[ 0.4213 \times 100 \approx 42.1\% \] **Final Answer:** \[ \boxed{42.1\%} \] *(rounded to the nearest tenth)*

To calculate the probability of winning the teddy bear, let's first find the probability of losing the game. The probability of missing a bottle in one try is 5 out of 6, or 5/6, since there are 6 bottles. The probability of missing all three tries is the multiplication of these individual probabilities: (5/6) * (5/6) * (5/6) = 125/216. Now, let's find the probability of winning. The winning probability is the complement of losing probability: 1 - 125/216 = 91/216. Now, convert the winning probability into percentage: (91/216) * 100 ≈ 42.1% So, the probability of winning the teddy bear is approximately 42.1%.