Questions: A Web site reported in 2018 that 47% of all Web traffic comes from laptops and desktops. Complete parts a through d below based on if a random sample of 11 households was selected. a. What is the probability that exactly two households used a laptop or desktop to access the web? The probability is 0.0401 . (Round to four decimal places as needed.) b. What is the probability that less than three households used a laptop or desktop to access the web? The probability is (Round to four decimal places as needed.)

Solution Steps

To find the probability that exactly two households used a laptop or desktop to access the web, we use the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 11 \) (number of trials),
• \( x = 2 \) (number of successes),
• \( p = 0.47 \) (probability of success),
• \( q = 1 - p = 0.53 \) (probability of failure).

Calculating this gives:

\[ P(X = 2) = \binom{11}{2} \cdot (0.47)^2 \cdot (0.53)^{9} = 0.0401 \]

Thus, the probability that exactly two households used a laptop or desktop is:

\[ \boxed{0.0401} \]

To find the probability that less than three households used a laptop or desktop, we sum the probabilities for \( x = 0, 1, \) and \( 2 \):

\[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \]

Calculating each term:

1. For \( x = 0 \): \[ P(X = 0) = \binom{11}{0} \cdot (0.47)^0 \cdot (0.53)^{11} = 0.0009 \]

2. For \( x = 1 \): \[ P(X = 1) = \binom{11}{1} \cdot (0.47)^1 \cdot (0.53)^{10} = 0.009 \]

3. For \( x = 2 \): \[ P(X = 2) = 0.0401 \quad \text{(as calculated previously)} \]

Now summing these probabilities:

\[ P(X < 3) = 0.0009 + 0.009 + 0.0401 = 0.049999999999999996 \]

Thus, the probability that less than three households used a laptop or desktop is approximately:

\[ \boxed{0.0500} \]

Final Answer