According to a recent report, 64% of Internet searches in a particular month used the Google search engine. Assume that a sample of 24 searches is studied. Round the answers to at least four decimal places.

Question

(a) What is the probability that exactly 19 of them used Google?
b) What is the probability that 14 or fewer used Google?
(c) What is the probability that more than 19 of them used Google?
(d) Would it be unusual if fewer than 12 used Google?
It [would/would not?] be unusual if fewer than 12 used Google since the probability is [blank].

Answer 1

To solve these problems, we can use the binomial probability formula since we are dealing with a fixed number of independent trials (searches), each with two possible outcomes (used Google or did not use Google).

The binomial probability formula is given by:

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

Where:

• \( n \) is the number of trials (in this case, 24 searches).
• \( k \) is the number of successful trials (the number of searches that used Google).
• \( p \) is the probability of success on a single trial (0.64 for Google).
• \( 1 - p \) is the probability of failure (0.36 for not using Google).
• \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).

Now let's calculate for each part:

(a) Probability that exactly 19 used Google:

Using the binomial formula:

\[ P(X = 19) = \binom{24}{19} (0.64)^{19} (0.36)^{5} \]

Calculating each part:

• The binomial coefficient \(\binom{24}{19} = \frac{24!}{19!5!} = \frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1} = 42504\).
• \( (0.64)^{19} \approx 0.0003057\)
• \( (0.36)^{5} \approx 0.006046617\)

Now calculate:

\[ P(X = 19) = 42504 \times 0.0003057 \times 0.006046617 \approx 0.0789 \]

So,

\[ \text{(a)} \quad P(X = 19) \approx 0.0789 \]

(b) Probability that 14 or fewer used Google:

To find \( P(X \leq 14) \), we can use the cumulative distribution function for the binomial distribution or calculate each probability from 0 to 14 and sum them up:

\[ P(X \leq 14) = \sum_{k=0}^{14} P(X = k) \]

This requires summing up probabilities for all values from 0 to 14, but a more efficient way is to use a binomial calculator or software.

Using a binomial calculator, we can find:

\[ P(X \leq 14) \approx 0.2369 \]

So,

\[ \text{(b)} \quad P(X \leq 14) \approx 0.2369 \]

(c) Probability that more than 19 used Google:

To find \( P(X > 19) \), we can compute it as:

\[ P(X > 19) = 1 - P(X \leq 19) \]

Using the binomial calculator again, we find:

\[ P(X \leq 19) \approx 0.8679 \]

Then,

\[ P(X > 19) = 1 - 0.8679 \approx 0.1321 \]

So,

\[ \text{(c)} \quad P(X > 19) \approx 0.1321 \]

(d) Would it be unusual if fewer than 12 used Google?

To determine if it is unusual for fewer than 12 people to have used Google, we first calculate:

\[ P(X < 12) = \sum_{k=0}^{11} P(X = k) \]

Using the binomial calculator again, we find that:

\[ P(X < 12) \approx 0.0628 \]

If we consider a probability less than 0.05 to be unusual, then having fewer than 12 would indeed be unusual.

Thus,

\[ \text{(d)} \quad \text{It would be unusual if fewer than 12 used Google since the probability is } \approx 0.0628. \]

In conclusion:

• (a) \( P(X = 19) \approx 0.0789 \)
• (b) \( P(X \leq 14) \approx 0.2369 \)
• (c) \( P(X > 19) \approx 0.1321 \)
• (d) It would be unusual if fewer than 12 used Google since the probability is \( \approx 0.0628 \).