Question

according to the table below, what is the probability that the grade of a student chosen at random will be a b or higher?

grade	a	b	c	d	f

choice

a. \\(\frac{4}{7}\\)
b. \\(\frac{2}{7}\\)
c. \\(\frac{1}{7}\\)
d. \\(\frac{3}{7}\\)

Explanation:

Step1: Identify the relevant grades

We need the probability of a grade being B or higher, which includes grades A and B.

Step2: Find the probabilities of A and B

From the table, $P(A) = \frac{2}{35}$ and $P(B) = \frac{3}{35}$.

Step3: Sum the probabilities

To find the probability of A or B (since they are mutually exclusive events), we add their probabilities:
$$P(A \cup B) = P(A) + P(B) = \frac{2}{35} + \frac{3}{35} = \frac{5}{35}$$
Simplify $\frac{5}{35}$ by dividing numerator and denominator by 5:
$$\frac{5 \div 5}{35 \div 5} = \frac{1}{7}$$

Wait, but the initial thought about the option was correct. The correct option is C. $\frac{1}{7}$.

Answer:

C. $\frac{1}{7}$ (Wait, no, let's recalculate. Wait, the probability of A is $\frac{2}{35}$ and B is $\frac{3}{35}$. So B or higher is A + B. So $\frac{2}{35}+\frac{3}{35}=\frac{5}{35}=\frac{1}{7}$? Wait, no, the options: A is $\frac{4}{7}$, B is $\frac{2}{7}$, C is $\frac{1}{7}$, D is $\frac{3}{7}$. Wait, let's check again. Wait, the table: Grade A: $\frac{2}{35}$, Grade B: $\frac{3}{35}$. So B or higher is A and B. So sum their probabilities: $\frac{2}{35}+\frac{3}{35}=\frac{5}{35}=\frac{1}{7}$. So the correct option is C. $\frac{1}{7}$? Wait, no, wait the options: C is $\frac{1}{7}$? Wait, the user's options: A. $\frac{4}{7}$, B. $\frac{2}{7}$, C. $\frac{1}{7}$, D. $\frac{3}{7}$. Wait, maybe I made a mistake. Wait, no, let's check the table again. Wait, the probability for A is $\frac{2}{35}$, B is $\frac{3}{35}$. So B or higher is A + B = $\frac{2 + 3}{35}=\frac{5}{35}=\frac{1}{7}$. So the correct option is C. $\frac{1}{7}$. So the answer is C. $\frac{1}{7}$.