To compute the probability, we can use the complementary probability, which means finding the probability that we do not get at least two heads and subtracting it from 1.
The sample space for three coin flips is:
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
There are two ways in which we don’t get at least 2 heads:
1. Flip no heads (all tails): TTT
2. Flip only one head: HTT, THT, TTH
Now, let’s compute the probability of each scenario. Since each coin flip is independent and has two possible outcomes (head or tail) with equal probability, we get:
1. No heads: $P(TTT) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
2. One head: $P(HTT) = P(THT) = P(TTH) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
Since there are three ways of flipping just one head, we have:
$$P(\text{not at least 2 heads}) = P(TTT) + 3 P(\text{one head}) = \frac{1}{8} + 3 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$$
Hence, the probability of getting at least 2 heads is given by the complementary probability:
$$P(\text{at least 2 heads}) = 1 - P(\text{not at least 2 heads}) = 1 - \frac{1}{2} = \frac{1}{2}$$
Therefore, the probability of getting at least two heads when flipping all three coins simultaneously is $\frac{1}{2}$.