Question 1171385: a family is planning an 8-hr road trip and is going to take along a porrtable video game system for the kids to play. The father is worried that the batteries will go dead before the end of the trip so he takes along an extra set. From experience he know the batteries will typically need replaced once every 10 hr. Assuming that the number of times the batteries need to be replaced in 1 hour has a poisson distribution and that the kids play the video game for all 8 hours, find the probability that the two sets of batteries last the entire trip.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $X$ be the number of times the batteries need to be replaced in 10 hours.
We are given that the batteries typically need to be replaced once every 10 hours.
So, the average rate of replacement is $\lambda = \frac{1}{10}$ replacements per hour.
We are interested in an 8-hour period, so the average number of replacements in 8 hours is:
$$ \lambda_{8} = \lambda \times 8 = \frac{1}{10} \times 8 = \frac{8}{10} = 0.8 $$
We are assuming the number of replacements follows a Poisson distribution.
Let $Y$ be the number of times the batteries need to be replaced in 8 hours. Then $Y \sim Poisson(\lambda_{8})$, where $\lambda_{8} = 0.8$.
The probability mass function of a Poisson distribution is given by:
$$ P(Y = k) = \frac{e^{-\lambda_{8}} \lambda_{8}^k}{k!} $$
The father takes along two sets of batteries. This means they can replace the batteries twice.
We want to find the probability that the batteries last the entire trip, which means the number of replacements must be 0, 1, or 2.
So, we want to find $P(Y \le 2) = P(Y = 0) + P(Y = 1) + P(Y = 2)$.
$$ P(Y = 0) = \frac{e^{-0.8} (0.8)^0}{0!} = e^{-0.8} \approx 0.4493 $$
$$ P(Y = 1) = \frac{e^{-0.8} (0.8)^1}{1!} = 0.8 e^{-0.8} \approx 0.8 \times 0.4493 \approx 0.3594 $$
$$ P(Y = 2) = \frac{e^{-0.8} (0.8)^2}{2!} = \frac{0.64 e^{-0.8}}{2} = 0.32 e^{-0.8} \approx 0.32 \times 0.4493 \approx 0.1438 $$
Now, we sum these probabilities:
$$ P(Y \le 2) = P(Y = 0) + P(Y = 1) + P(Y = 2) \approx 0.4493 + 0.3594 + 0.1438 \approx 0.9525 $$
Therefore, the probability that the two sets of batteries last the entire trip is approximately 0.9525.
Final Answer: The final answer is $\boxed{0.9525}$
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