Question

A box contains 4 red, 5 blue and 1 green marble. A child randomly takes out a marble from the box, notes down the colour and puts it back in the box. If the activity is repeated 3 times, what is the probability that at least one marble is red ?

• A. \(\frac{27}{125}\)
• B. \(\frac{8}{125}\)
• C. \(\frac{2}{125}\)
• D. \(\frac{98}{125}\)

Hint:

"At least one" problems are almost always solved faster using the complement (1 - none).
Replacement means the denominator stays constant for every draw.

Answer 1

The Correct Option is D

Concept:
• This is a case of sampling with replacement, making each draw an independent Bernoulli trial.
• Total number of marbles \(= 4 (\text{red}) + 5 (\text{blue}) + 1 (\text{green}) = 10\).
• Complementary probability: \(P(\text{at least one red}) = 1 - P(\text{no red marbles})\).

Step 1: Find the probability of drawing a red marble in a single trial
\[ P(\text{Red}) = \frac{\text{Number of red marbles}}{\text{Total marbles}} = \frac{4}{10} = \frac{2}{5} \]

Step 2: Find the probability of NOT drawing a red marble in a single trial
\[ P(\text{Not Red}) = 1 - P(\text{Red}) = 1 - \frac{2}{5} = \frac{3}{5} \]

Step 3: Calculate the probability of drawing no red marbles in 3 independent trials
Since the events are independent (due to replacement):
\[ P(\text{No red in 3 trials}) = (P(\text{Not Red}))^3 = \left(\frac{3}{5}\right)^3 = \frac{27}{125} \]

Step 4: Calculate the final probability of at least one red marble
Using the complementary principle:
\[ P(\text{At least one red}) = 1 - P(\text{No red in 3 trials}) \] \[ = 1 - \frac{27}{125} = \frac{125 - 27}{125} = \frac{98}{125} \]