Math Snap

STEP 1

1. The battery life follows an exponential distribution with a rate parameter $\lambda = \frac{1}{365}$.
2. The exponential distribution probability density function is given by $f(x; \lambda) = \lambda e^{-\lambda x}$.
3. The cumulative distribution function (CDF) for an exponential distribution is $F(x; \lambda) = 1 - e^{-\lambda x}$.
4. The probability that a battery lasts beyond a certain time is the complement of the CDF.

STEP 2

1. Calculate the probability that a battery will fail before 180 days.
2. Calculate the probability that a battery will last beyond 1 year.
3. Calculate the probability that both batteries in a device last beyond 1 year.

STEP 3

Calculate the probability that a battery will fail before 180 days using the CDF:
$$F(180; \lambda) = 1 - e^{-\frac{1}{365} \times 180}$$

STEP 4

Compute the value:
$$F(180; \lambda) = 1 - e^{-\frac{180}{365}}$$ $$F(180; \lambda) = 1 - e^{-0.49315}$$ $$F(180; \lambda) \approx 1 - 0.6107$$ $$F(180; \lambda) \approx 0.3893$$ The probability that an alkaline battery will fail before 180 days is approximately $\boxed{0.3893}$.

STEP 5

Calculate the probability that a battery will last beyond 1 year using the complement of the CDF:
$$P(X > 365) = 1 - F(365; \lambda) = e^{-\frac{1}{365} \times 365}$$

STEP 6

Compute the value:
$$P(X > 365) = e^{-1}$$ $$P(X > 365) \approx 0.3679$$ The probability that an alkaline battery will last beyond 1 year is approximately $\boxed{0.3679}$.

STEP 7

Calculate the probability that both batteries last beyond 1 year. Since the batteries are independent, the probability is the product of the individual probabilities:
$$P(\text{Both last} > 365) = P(X_1 > 365) \times P(X_2 > 365)$$ $$P(\text{Both last} > 365) = (0.3679)^2$$

Compute the value:
$$P(\text{Both last} > 365) = 0.3679 \times 0.3679$$ $$P(\text{Both last} > 365) \approx 0.1353$$ The probability that both batteries last beyond 1 year is approximately $\boxed{0.1353}$.