Question

High School Grades of First-Year College Students Forty-seven percent of first-year college students enrolled in 2005 had an average grade of A in highschool compared to 20% of first-year college students in 1970. Choose 6 first-year college students at random enrolled in 2005. Find the probability that

a. All had an A average in high school
b. None had an A average in high school
c. At least 1 had an A average in high school​

Answer 1

a. Probability all 6 first-year college students had an A average in high school ≈ 4.71%.

b. Probability none of the 6 had an A average ≈ 0.58%.

c. Probability at least 1 had an A average ≈ 99.42%.

To find the probabilities, we can use the binomial probability formula, which is given by:

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

where:

-
`\( n \)` is the number of trials (in this case, the number of first-year college students chosen),

-
`\( k \)` is the number of successes (in this case, the number of students with an A average in high school),

-
`\( p \)` is the probability of success on a single trial.

For this problem:

-
`\( n = 6 \)` (6 first-year college students chosen),

-
`\( p = 0.47 \)` (probability of having an A average in high school).

Let's calculate the probabilities:

a. Probability that all had an A average in high school:

`\[ P(\text{all A}) = \binom{6}{6} \cdot (0.47)^6 \cdot (1 - 0.47)^(0) \]`

b. Probability that none had an A average in high school:

`\[ P(\text{none A}) = \binom{6}{0} \cdot (0.47)^0 \cdot (1 - 0.47)^(6) \]`

c. Probability that at least 1 had an A average in high school:

`\[ P(\text{at least 1 A}) = 1 - P(\text{none A}) \]`

Let's calculate these probabilities:

a. Probability that all had an A average in high school:

`\[ P(\text{all A}) = \binom{6}{6} \cdot (0.47)^6 \cdot (1 - 0.47)^0 \]`

`\[ P(\text{all A}) = 1 \cdot 0.0471441 \cdot 1 \]`

`\[ P(\text{all A}) \approx 0.0471441 \]`

b. Probability that none had an A average in high school:

`\[ P(\text{none A}) = \binom{6}{0} \cdot (0.47)^0 \cdot (1 - 0.47)^6 \]`

`\[ P(\text{none A}) = 1 \cdot 1 \cdot (1 - 0.47)^6 \]`

`\[ P(\text{none A}) \approx 0.0057941 \]`

c. Probability that at least 1 had an A average in high school:

`\[ P(\text{at least 1 A}) = 1 - P(\text{none A}) \]`

`\[ P(\text{at least 1 A}) \approx 1 - 0.0057941 \]`

`\[ P(\text{at least 1 A}) \approx 0.9942059 \]`

So, the probabilities are:

a.
`\( \approx 0.0471441 \)`

b.
`\( \approx 0.0057941 \)`

c.
`\( \approx 0.9942059 \)`