Question

There are 6 tickets to the theater, 4 of which are for seats in the front row. If 3 tickets are selected at random, what is the probability that exactly two of them are for the front row?

• A. 0.6
• B. 0.7
• C. 0.9
• D. $\dfrac{1}{3}$

Hint:

For “exactly k successes” without order, use combinations:
\(\dfrac{\binom{\text{success pool}}{k} \; \binom{\text{failure pool}}{n-k}}{\binom{\text{total}}{n}}\).

Answer 1

The Correct Option is A

Step 1: Identify favorable selection pattern.
We want exactly 2 front-row tickets and 1 non-front ticket.
Front-row (F) tickets $=4$, Non-front (N) tickets $=2$.
Step 2: Count favorable ways.
Choose 2 from 4 F tickets: $\binom{4}{2} = 6$.
Choose 1 from 2 N tickets: $\binom{2}{1} = 2$.
Favorable outcomes $= 6 \times 2 = 12$.
Step 3: Count total ways.
Total ways to choose any 3 from 6 tickets: $\binom{6}{3}=20$.
Step 4: Compute probability.
$P=\dfrac{12}{20}=\dfrac{3}{5}=0.6$.
\[ \boxed{0.6} \]