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I did a quick search, couldn't find an answer that helps me understand this question:

Supposed there is a company with a problem, they can either assign one big team C (30% chance to solve) or two smaller teams A (20% chance) and B (10% chance) to solve it. If team A solves the problem - team B has a 5% chance to solve the problem. Which team setup is better?

Initial attempt:

P(A) = 0.20, P('A) = 0.80, P(A|B) = 0.10, P('A|B) = 0.90, P(A|'B)= 0.11, P('A|'B) = 0.7888 P(B) = 0.10, P('B) = 0.90, P(B|A) = 0.05, P('B|A) = 0.95, P(B|'A)=0.1125, P('B|'A) = 0.8875

This is where I got stuck and have a few questions:

1) For simultaneous events, with conditional probability, does the sequence of events lead to different probabilities? As in does Team A succeeding first or not need to be taken into account as well as Team B - leading to total probability of solving the problem = P(B|A)P(A) + P('B|A)P(A) + P(A|B)P(B) + P('A|B)P(B)?

2) If we assume that the team that completes it first will notify the other team, how would we be able to include or combine the earlier probability in? That is if A finishes it first (P(A) = 0.20) + B finishes it first (P(B) = 0.10) + A doesn't finish it, and B does plus the opposite (P('A AND B) + P(A AND 'B))?

I think I am thinking too much into this problem... would sincerely appreciate your help as I honestly suck at probability questions and would love to learn how to conceptualize these.

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Don't get hung up on which team finishes it "first". This question says nothing about causality, all it does is give you a hint towards what you really want, which is the probability that any team solves it, whether it's $A$, $B$, or both. The question gives you $P(A)$, $P(B)$, and $P(B \mid A)$. The influence of one team finishing "first" on the other team can be due to many complex causal effects, outside the scope of this problem, and none of which are important for the quantity you want.

As the manager, you only care about the quantity:

$$ P(A \lor B) = P(A) + P(B) - P(A,B). $$

Note the identity,

$$ P(B \mid A) = \frac{P(A,B)}{P(A)}. $$

With that, and the knowledge given by the question, you have enough to figure out what you, as the manager, care about.

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  • $\begingroup$ Thank you very much! I was really just thinking too much into this... $\endgroup$ – ConfusedProbability Feb 4 at 13:36

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