Question

A box has 4 green, 8 blue, and 3 red pens. A student picks up a pen at random, checks its color and replaces it in the box. He repeats this process 3 times. The probability that at least one pen picked was red is:

• A. 124/125
• B. 125/125
• C. 64/125
• D. 125/125

Hint:

When finding the probability of "at least one" event, calculate the complement (probability of none) and subtract from 1.

Answer 1

The Correct Option is C

We are given a box with 4 green, 8 blue, and 3 red pens. The total number of pens is: \[ 4 + 8 + 3 = 15. \] The probability of picking a red pen in one trial is: \[ P(\text{red}) = \frac{3}{15} = \frac{1}{5}. \] The probability of not picking a red pen in one trial is: \[ P(\text{not red}) = 1 - P(\text{red}) = 1 - \frac{1}{5} = \frac{4}{5}. \] The probability of not picking a red pen in 3 trials is: \[ \left(\frac{4}{5}\right)^3 = \frac{64}{125}. \] Therefore, the probability of picking at least one red pen is: \[ 1 - \frac{64}{125} = \frac{61}{125}. \] So, the correct answer is (3).