There are 3 lucky tickets among 10 lottery tickets. 3 tickets are
drawn at random. What is a probability that one ticket selected randomly
from those three is lucky?
For exactly one lucky ticket in three draws without replacement:
3/10 * 7/9 * 6/8 = ?
Do we need to use A or C (combainations)
To find the probability that one ticket selected randomly from the three drawn tickets is lucky, we need to calculate the ratio of favorable outcomes to total outcomes.
In this case, there are 3 lucky tickets out of a total of 10 tickets. Since we are drawing 3 tickets out of the 10, we can use the concept of combinations.
The total number of ways to select 3 tickets out of 10 is given by the combination formula, which is denoted as C(10, 3) and can be calculated as:
C(10, 3) = 10! / (3! * (10-3)!)
= 10! / (3! * 7!)
= (10 * 9 * 8) / (3 * 2 * 1)
= 120
Now, let's consider the favorable outcomes, which is the number of ways to select 1 lucky ticket from the 3 drawn lucky tickets. This can be represented as C(3, 1) and calculated as:
C(3, 1) = 3! / (1! * (3-1)!)
= 3! / (1! * 2!)
= (3 * 2) / (1 * 1)
= 6 / 2
= 3
Therefore, the probability that one ticket selected randomly from the three drawn tickets is lucky is given by:
Probability = favorable outcomes / total outcomes
= C(3, 1) / C(10, 3)
= 3 / 120
= 1 / 40
= 0.025
So, the probability is 0.025 or 2.5%.