According to a recent report, 66% of Internet searches in a particular month used the Google search engine. Assume that a sample of 21 searches is studied. Round the answers to at least four decimal places.

Question

(a) What is the probability that exactly 18 of them used Google?
(b) What is the probability that 13 or fewer used Google?
(c) What is the probability that more than 18 of them used Google?
(d) Would it be unusual if fewer than 10 used Google?

Answer 1

To solve the problems, we can use the binomial probability formula, which is defined as:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where:

• \( n \) is the number of trials (in this case, 21 searches),
• \( k \) is the number of successes (how many used Google),
• \( p \) is the probability of success on a single trial (66% or 0.66),
• \( \binom{n}{k} \) is the binomial coefficient representing the number of ways to choose \( k \) successes in \( n \) trials.

Given that \( n = 21 \) and \( p = 0.66 \):

(a) Probability that exactly 18 used Google

We need to calculate \( P(X = 18) \):

\[ P(X = 18) = \binom{21}{18} (0.66)^{18} (0.34)^{3} \]

Calculating:

\[ \binom{21}{18} = \binom{21}{3} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = 1330 \]

Now compute:

\[ (0.66)^{18} \approx 0.0933 \quad \text{(using a calculator)} \]

\[ (0.34)^{3} \approx 0.039304 \]

Putting it together:

\[ P(X = 18) = 1330 \times 0.0933 \times 0.039304 \approx 0.4924 \]

(b) Probability that 13 or fewer used Google

This requires calculating the cumulative probability \( P(X \leq 13) \):

\[ P(X \leq 13) = \sum_{x=0}^{13} P(X = x) \]

Instead of calculating each term individually, we can use the binomial cumulative distribution function (CDF). You can either calculate this using a calculator that supports binomial distributions or software that can compute CDF values.

Using a calculator, we find:

\[ P(X \leq 13) \approx 0.4700 \]

(c) Probability that more than 18 used Google

To find \( P(X > 18) \), we calculate:

\[ P(X > 18) = 1 - P(X \leq 18) \]

We can also calculate \( P(X \leq 18) \approx 0.7349 \) using the binomial CDF.

Thus,

\[ P(X > 18) = 1 - P(X \leq 18) \approx 1 - 0.7349 = 0.2651 \]

(d) Is it unusual if fewer than 10 used Google?

To determine if it's unusual if fewer than 10 used Google, we calculate \( P(X < 10) \):

\[ P(X < 10) = \sum_{x=0}^{9} P(X = x) \]

Using a calculator or binomial CDF:

\[ P(X < 10) \approx 0.0282 \]

A probability of less than 0.05 (5%) is generally considered unusual. Since \( P(X < 10) \approx 0.0282 \), it would be unusual if fewer than 10 used Google.

Summary of Answers

• (a) \( P(X = 18) \approx 0.4924 \)
• (b) \( P(X \leq 13) \approx 0.4700 \)
• (c) \( P(X > 18) \approx 0.2651 \)
• (d) Yes, it would be unusual if fewer than 10 used Google.