There are two ways of solving this problem. Let's go with the easier one.
Note that there are several ways you could have at least one person be left-handed out of 6 people: 1, 2, 3, 4, 5, or all 6 people are left handed. However let's look at the opposite. How many ways are there for nobody to be left-handed? Just one way: every person is right-handed.
So, we calculate the probability of getting 6 random people that are all right-handed.
*Note that each person is independent of the next, so we multiply the probabilities.
P(1 random person is right-handed)
= 100% - P(1 random person is left-handed)
= 100% - 18%
= 82%
P(6 random people are all right-handed)
= P(1 random person is right-handed)^6
= (82%)^6
= 0.3040066714
So, this situation is the exact opposite of what we are looking for. Thus:
P(at least 1 person out of 6 random people is left-handed)
= 100% - P(6 random people are all right-handed)
= 1 - 0.3040066714
= 0.6959933286
So, the probability of at least 1 person among 6 randomly chosen people is left-handed is about 69.6%
