Question

A basket contains 4 red, 5 blue and 3 green marbles. If 3 marbles are picked at random, what is the probability that either all are green or all are red?

• A. $\frac{7}{44}$
• B. $\frac{7}{12}$
• C. $\frac{5}{12}$
• D. $\frac{1}{44}$

Hint:

Use combinations when order does not matter: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$

Answer 1

The Correct Option is A

Concept: \[ \text{Probability} = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} \]

Step 1: Total marbles.
\[ 4 + 5 + 3 = 12 \]

Step 2: Total ways to pick 3 marbles.
\[ \binom{12}{3} = 220 \]

Step 3: Favourable cases.

• All green: $\binom{3}{3} = 1$
• All red: $\binom{4}{3} = 4$ Total favourable: \[ 1 + 4 = 5 \]

Step 4: Probability.
\[ \frac{5}{220} = \frac{1}{44} \]

Step 5: Final conclusion.
Thus, probability is: \[ $\frac{1{44}$} \]