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I was wondering if my answer to this question is correct.

A, B and C live in an apartment together. A gets up in the morning before B and C is up last. First, to the kitchen, A encounters a sink full of dirty dishes and washes them with probability $pA=1/2$. Similarly, after A has left, on encountering dirty dishes, B washes them with probability $pB=1/3$. Likewise, C washes them with probability $pC=1/4$.

a) What is the probability that the dishes get done?

b) If the dishes get done, what is the probability that C washed them?

A) P[dishes are done] = $(1/2) * (1/3) * (1/4) = 1/24$

B) Use conditional probability

P[C washed dishes] = P[dishes are done] * P[A washed dishes] * P[Bwashed dishes]

P[C washed dishes] = $(1/24) * (1/2) * (1/3) = 1/144$

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    $\begingroup$ A) How can the probablility that the dishes are done at all be less than the probability that they are done by A? Imagine A gets up and does the dishes with the probability of 1/2 and then leaves. Then image a meteor hits the apartment and everyone dies. The probability that the dishes were done is 1/2. Now imagine that the meteor didn't hit and B and C wake up. You are saying the probabilty that the dishes getting done without B or C possibly doing them, is greater than if B and C might actually do them? That's not possible, is it? $\endgroup$
    – fleablood
    Sep 18, 2017 at 2:25
  • $\begingroup$ Does that mean that the probability of all the dishes getting washed is the sum of all the individual probabilities? $\endgroup$ Sep 18, 2017 at 2:36
  • $\begingroup$ If everyone dies, the probability still exists in space? I am also not sure a person washes dirty dishes either partially or fully. $\endgroup$ Sep 18, 2017 at 2:37
  • $\begingroup$ "Does that mean that the probability of all the dishes getting washed is the sum of all the individual probabilities?" No. I'm just pointing out that if $A$ has a $1/2$ probability then $A$ or $B$ or $C$ must have a greater than 1/2 probability. So getting $1/24$ is clearly not correct. As for what is correct. Well, we can't just guess. mulitply? No? Add? No? Take the exponents? No? $\endgroup$
    – fleablood
    Sep 18, 2017 at 3:39

2 Answers 2

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a) What is the probability that the dishes get done?

1 - "Probability that the dishes does not get washed by anyone"

and

"Probability that the dishes does not get washed by anyone" = (1 - pA)(1-pB)(1-pC)

b) If the dishes get done, what is the probability that C washed them?

(1-pA)*(1-pB)*pC / ( pA + (1-pA)pB + (1-pA)(1-pB)*pC )

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In addition to the comment of guest, giving a formula, the answer works out to be $\frac34$ and $\frac19$, in that order. (Also, the denominator of his second formula simplifies to $1-(1-a)(1-b)(1-c)$.)

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