# Given Information - There are **12 sensors** tested simultaneously. - Each sensor fails independently with probability **.1**. - Let \( X \) = number of sensors that fail during the test. # What We Have to Find - Probability that **at most two sensors fail** during the testing period, i.e., \( P(X \leq 2) \). # Definition or Concept Used - This is a **binomial probability problem**. - The random variable \( X \) (number of failing sensors) follows a **Binomial distribution**: \[ X \sim \text{Binomial}(n=12, p=.1) \] - The probability mass function for the binomial distribution: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] # Step-by-Step Solution Calculate \( P(X \leq 2) = P(X=) + P(X=1) + P(X=2) \). \[ P(X=) = \binom{12}{} (.1)^ (.9)^{12} = 1 \times 1 \times .9^{12} \] \[ P(X=) = .2824 \] \[ P(X=1) = \binom{12}{1} (.1)^1 (.9)^{11} = 12 \times .1 \times .9^{11} \] \[ P(X=1) = 12 \times .1 \times .3138 = .3766 \] \[ P(X=2) = \binom{12}{2} (.1)^2 (.9)^{10} = 66 \times .01 \times .3487 \] \[ P(X=2) = 66 \times .01 \times .3487 = .2301 \] Sum the probabilities: \[ P(X \leq 2) = .2824 + .3766 + .2301 = .8891 \] # Summary The probability that **at most two sensors fail during the testing period** is **.8891**.

# Given Information - Number of sensors tested: **12** - Probability of a sensor failing: **0.1** - Let \( X \) = number of sensors that fail during the testing period. # What We Have to Find - Probability that **at most two sensors fail** during the testing period, i.e., \( P(X \leq 2) \). # Definition or Concept Used - The scenario follows a **binomial probability distribution**: \[ X \sim \text{Binomial}(n=12, p=0.1) \] - The probability mass function for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] # Step-by-Step Solution Calculate \( P(X \leq 2) \) by finding \( P(X=0) \), \( P(X=1) \), and \( P(X=2) \): 1. **Calculate \( P(X=0) \)**: \[ P(X=0) = \binom{12}{0} (0.1)^0 (0.9)^{12} \] \[ P(X=0) = 1 \times 1 \times (0.9)^{12} \approx 0.2824 \] 2. **Calculate \( P(X=1) \)**: \[ P(X=1) = \binom{12}{1} (0.1)^1 (0.9)^{11} \] \[ P(X=1) = 12 \times 0.1 \times (0.9)^{11} \approx 12 \times 0.1 \times 0.3138 \approx 0.3766 \] 3. **Calculate \( P(X=2) \)**: \[ P(X=2) = \binom{12}{2} (0.1)^2 (0.9)^{10} \] \[ P(X=2) = 66 \times 0.01 \times (0.9)^{10} \approx 66 \times 0.01 \times 0.3487 \approx 0.2301 \] 4. **Sum the probabilities**: \[ P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) \approx 0.2824 + 0.3766 + 0.2301 = 0.8891 \] # Summary The probability that **at most two sensors fail during the testing period** is **0.8891**.