Question

Two guys walk into a bar; the third one ducks. If these three men acted independently, and the probability is 0.7 that a person who is about to walk into a bar instead ducks to avoid the bar, then what was the probability of observing the first two walking into the bar and the third ducking?

Answer 1

Answer:

$0.063$

Explanation:

Given

Probability of a person to not enter into a bar or ducking   is $0.7$

Probability of a person to  enter into a bar  

$1 -$ (Probability of a person to not enter into a bar or ducking)

Substituting the given value, we get

Probability of a person to enter into a bar  

$= 1 - 0.7 \\= 0.3$

Total three men attempts to enter into the bar and their course of action is independent of each others

Thus, probability of observing the first two walking into the bar and the third ducking will be equal to the product of individual probabilities

$= 0.7 * 0.3 * 0.3\\= 0.063$