Assume that we observe three draws, X1,X2,X3 from a Bernoulli distribution with parameter p=12. For example, imagine that in the model for the preferred head direction for kissing, either direction were actually equally likely and we observed three kissing couples.
What is the probability of observing at least two ones, i.e., what is P(∑3i=1Xi≥2)?
0.5
To find the probability of observing at least two ones, let's calculate the probability of each possible outcome and add them up.
The random variables X1, X2, and X3 are drawn from a Bernoulli distribution with parameter p=1/2 (since you mentioned that either direction of kissing is equally likely).
To calculate P(∑3i=1Xi≥2), we need to take into account three possible outcomes:
1. All three draws are ones (1, 1, 1).
2. Two of the three draws are ones (1, 1, 0 or 1, 0, 1 or 0, 1, 1).
3. At least two of the three draws are ones (1, 1, 1 or 1, 1, 0 or 1, 0, 1 or 0, 1, 1).
Now, let's calculate each probability:
1. Probability of getting three ones: P(X1 = 1) * P(X2 = 1) * P(X3 = 1) = (1/2) * (1/2) * (1/2) = 1/8.
2. Probability of getting two ones: P(X1 = 1) * P(X2 = 1) * P(X3 = 0) + P(X1 = 1) * P(X2 = 0) * P(X3 = 1) + P(X1 = 0) * P(X2 = 1) * P(X3 = 1) = (1/2) * (1/2) * (1/2) + (1/2) * (1/2) * (1/2) + (1/2) * (1/2) * (1/2) = 3/8.
3. Probability of at least two ones: P(≥2) = P(getting three ones) + P(getting two ones) = 1/8 + 3/8 = 4/8 = 1/2.
Therefore, the probability of observing at least two ones, P(∑3i=1Xi≥2), is 1/2 or 50%.