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Say I have three independent events with a probability of .25, .5, and .75 chance success for the individual events. How would I determine the probability that 0, 1, 2, or 3 of the events are successful?

Ben Mathews
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  • Can you say more about the structure of the situation? Eg, since >1 event could be successful w/i this scheme, there must be >1 *opportunity* for events to be successful. Consider a Binomial, where we say there are $n$ *trials* (eg, coin flips). How many such opportunities / trials are there w/i your situation? Is it possible for >1 event to be successful on the same trial? Etc. – gung - Reinstate Monica Sep 25 '13 at 14:01
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    Is this homework? Then it should be flagged as such. My answer below assumes that it is and therefore does not spell out the answer. – Erik Sep 25 '13 at 14:15
  • Ponder this: $(3/4+x/4)(2/4+2x/4)(1/4+3x/4)$ = $\frac{3}{32}x^0+\frac{13}{32}x^1+\frac{13}{32}x^2+\frac{3}{32}x^3.$ This is the method explained at http://stats.stackexchange.com/a/5482/919. – whuber Sep 25 '13 at 14:29
  • No, not homework. Just trying to figure out a problem and my stats class was way too long ago. I remember my basic probability including the answer below. I was just hoping that there was a formula that would simplify the calculations. – Ben Mathews Sep 25 '13 at 18:46

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You just need to apply three rules:

  1. If A and B are independent $$ P(A \cap B) = P(A)P(B)$$

  2. If $X = A \cup B$ then

    $$P(X) = P(A)+P(B)-P(A \cap B)$$

$$P(A ^\complement) = 1- P(A)$$

and try to separate your statements into disjoints possibilities where you state exactly what happens with your three Events A,B, C. For example, zero successful events means exactly NOT A, NOT B and NOT C. One successful event means (A AND NOT B AND NOT C) OR (NOT A AND B AND NOT C) OR (NOT A AND NOT B AND C).

Remember that AND becomes $\cap$ when translating to sets and OR becomes $\cup$.

Erik
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