Question

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

• A. 0.6
• B. 0.7
• C. 0.9
• D. \(\frac{1}{3}\)

Answer 1

The Correct Option is A

To solve the problem of determining the probability that two out of three randomly selected tickets are for the front row, we will use the method of combinations.

First, identify the relevant combinations:

1. Total number of tickets = 6
2. Number of front-row tickets = 4
3. Number of other tickets = 6 - 4 = 2
4. Total tickets selected = 3

We need the probability that 2 tickets are from the front row and 1 ticket from the other rows. Calculate the combinations:

1. Ways to choose 2 front-row tickets: \( \binom{4}{2} \)
2. Ways to choose 1 other ticket: \( \binom{2}{1} \)

The number of favorable outcomes is the product of the two combinations above:

\(\binom{4}{2} \cdot \binom{2}{1} = 6 \cdot 2 = 12\)

Total ways to choose any 3 tickets from 6:

\(\binom{6}{3} = 20\)

Hence, the probability is:

\(P(\text{two front-row tickets}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{12}{20} = 0.6\)

Therefore, the probability that two of the selected tickets are for the front row is 0.6.