Speedometer Probability 7Aea43

1. **State the problem:** We have 14 speedometers total: 12 correctly calibrated and 2 not. We select 3 without replacement. The random variable $x$ is the number of not correctly calibrated speedometers selected. 2. **Find the total number of ways to select 3 speedometers from 14:** $$\text{Total ways} = \binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364$$ 3. **Calculate $P(x=0)$:** No not calibrated selected means all 3 are from the 12 correct ones. $$P(x=0) = \frac{\binom{12}{3} \times \binom{2}{0}}{\binom{14}{3}} = \frac{\binom{12}{3}}{364}$$ Calculate numerator: $$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$ So: $$P(x=0) = \frac{220}{364} \approx 0.6044$$ 4. **Calculate $P(x=1)$:** Exactly 1 not calibrated and 2 correct. $$P(x=1) = \frac{\binom{12}{2} \times \binom{2}{1}}{364}$$ Calculate numerator: $$\binom{12}{2} = \frac{12 \times 11}{2} = 66$$ $$\binom{2}{1} = 2$$ So: $$P(x=1) = \frac{66 \times 2}{364} = \frac{132}{364} \approx 0.3626$$ 5. **Calculate $P(x=2)$:** Exactly 2 not calibrated and 1 correct. $$P(x=2) = \frac{\binom{12}{1} \times \binom{2}{2}}{364}$$ Calculate numerator: $$\binom{12}{1} = 12$$ $$\binom{2}{2} = 1$$ So: $$P(x=2) = \frac{12 \times 1}{364} = \frac{12}{364} \approx 0.0330$$ 6. **Calculate $P(x=3)$:** Exactly 3 not calibrated. Since only 2 are not calibrated, this is impossible. $$P(x=3) = 0$$ 7. **Check probabilities sum to 1:** $$0.6044 + 0.3626 + 0.0330 + 0 = 1.0000$$ 8. **Find the mean (expected value) of $x$:** $$\mu = E(x) = \sum x P(x) = 0 \times 0.6044 + 1 \times 0.3626 + 2 \times 0.0330 + 3 \times 0 = 0 + 0.3626 + 0.0660 + 0 = 0.4286$$ **Final answers:** | $x$ | $P(x)$ | |-----|--------| | 0 | 0.6044 | | 1 | 0.3626 | | 2 | 0.0330 | | 3 | 0 | Mean = $0.4286$