Question

there are 4 red balls, 6 white balls, and 3 green balls in a bag. if one ball is drawn from the bag at random, what is the probability that it is not white?
a. \\(\frac{7}{13}\\)
b. \\(\frac{6}{13}\\)
c. \\(\frac{5}{6}\\)
d. \\(\frac{1}{7}\\)

Explanation:

Step1: Calculate total number of balls

Total balls = red + white + green = \( 4 + 6 + 3 = 13 \)? Wait, no, 4+6=10, 10+3=13? Wait, no, 4 red, 6 white, 3 green. So 4+6+3=13? Wait, but the options have denominators 13? Wait, no, the options have 13? Wait, A is 7/13? Wait, the original problem's options: A is 7/13? Wait, maybe I miscalculated. Wait, 4 red, 6 white, 3 green. So total balls: 4 + 6 + 3 = 13? Wait, no, 4+6=10, 10+3=13. Then, number of non - white balls: red + green = 4 + 3 = 7. So probability of non - white is number of non - white / total = 7/13. Wait, but the first option A is 7/13? Wait, maybe the original problem's options have a typo? Wait, the user's image: let me check again. Wait, the user's image shows options: A. 7/13, B. 6/13, C. 5/6, D. 1/7. Wait, maybe I misread. Let's re - do:

Total number of balls: \( 4+6 + 3=13 \)

Number of non - white balls: red (4) + green (3) = 7

Probability (not white) = \( \frac{\text{number of non - white balls}}{\text{total number of balls}}=\frac{4 + 3}{4+6 + 3}=\frac{7}{13} \)

Step2: Match with options

The option A is \( \frac{7}{13} \)

Answer:

A. \( \frac{7}{13} \)