Conditional Probability Help -- Is this a trick question?

[redsox123]

A conditional probability problem that I can't figure out: (PART 1) Tom is trying to figure out the probability that his memory of a ride in a fighter jet when he was a toddler is real or if it is just a product of his imagination. He grew up near an air force base -- the odds of a child in that area taking a ride on a fighter jet are 1/1000. Because he has no bias towards whether the event happened or not -- he is 50 percent sure it did, 50 percent sure it did not -- per Bayes's theory, there is a 1/1000 chance it did, the prior odds. (Is the assumption above true (y/n): Y(PART 2) If it is true, please answer the following: Of those that correctly recalled riding on the fighter jet (they actually did), 1/2 remembered spontaneously and 1/2 were actively thinking about their childhood. Of those that did not ride on the jet and merely imagined it, 1/100 remembered spontaneously, 99/100 were actively thinking about their childhood. If we were 100 percent sure that Tom remembered while actively thinking about his childhood, how does this affect the probability that he actually rode on the fighter jet? What if we were 75 percent certain? And 50 percent? And, finally, 25 percent? Is it impossible to calculate? To start, we can conclude that of 1000 children the following is true: RODE JET / REMEMBERED (1 child)......DIDN'T RIDE JET / IMAGINED (999 children)
a. Spontaneous Recall (1/2 * 1) = 1/2 child ...c. Spontaneous Recall (1/100 * 999) = 9.99 children
b. Active Recall (1/2 * 1) = 1/2 child..... d. Active Recall (99/100 * 999) = 989.01 children

Regarding 100 percent -- this seems straight-forward. If Tom is 100 percent sure he remembered while actively thinking about his childhood, then the odds of him riding on the jet would be expressed as b / d.And the probability would be unchanged for 50 percent -- this assumes we have no idea if the memory was spontaneously recalled or not, so the prior odds would remain. Regarding 25 and 75 percent: the only way I can think of solving is by averaging the values we got for 100 and 50 percent. What would be the formulaic way to solve? I thought I was pretty proficient at Bayes' theorem, I just can't wrap my head around this one. I don't know which prior probability I should start with.

 

[Valkarie]

To solve this problem using Bayes' theorem, we need to start with the prior odds and then adjust them based on the additional information. For 100% certainty:P(Rode Jet | Active Recall) = P(Active Recall | Rode Jet) * P(Rode Jet) / P(Active Recall) P(Rode Jet | Active Recall) = (1/2 * 1/1000) / (1/2 * 1 + 99/100 * 999)= 0.001 / 0.9995= 0.1%For 75% certainty:P(Rode Jet | Active Recall) = P(Active Recall | Rode Jet) * P(Rode Jet) / ((0.75 * P(Active Recall|Rode Jet)) + (0.25 * P(Active Recall|Did Not Ride Jet)))P(Rode Jet | Active Recall) = (1/2 * 1/1000) / ((0.75 * 1/2) + (0.25 * 99/100))= 0.001 / 0.5025= 2%For 50% certainty:P(Rode Jet | Active Recall) = P(Active Recall | Rode Jet) * P(Rode Jet) / ((0.5 * P(Active Recall|Rode Jet)) + (0.5 * P(Active Recall|Did Not Ride Jet)))P(Rode Jet | Active Recall) = (1/2 * 1/1000) / ((0.5 * 1/2) + (0.5 * 99/100))= 0.001 / 0.495= 0.2%For 25% certainty:P(Rode Jet | Active Recall) = P(Active Recall | Rode Jet) * P(Rode Jet) / ((0.25 * P(Active Recall|Rode Jet)) + (0.75 * P(Active Recall|Did Not Ride Jet)))P(Rode Jet | Active Recall) = (1/2 * 1/1000) / ((0.25 * 1/2) + (0.75 * 99/100))= 0.001 / 0.24875= 0.4%