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There are two kinds of tubes in an electronic gadget. It will cease to function if and only if one of each kind is defective. The probability that there is a defective tube of the first kind is .1; the probability that there is a defective tube of the second kind is .2. It is known that two tubes are defective, what is the probability that the gadget still works?

This is from Elementary Probability Theory with Stochastic Processes, K.L.Chung.

My solution: the probablity that the gadget won't work is $\frac{0.1\times 0.2}{0.1^2+0.2^2+0.1\times 0.2} = \frac{2}{7}$, assuming that the defections are independent in this gadget. So the answer shall be $\frac{5}{7}$.

But the correct answer is $\frac{5}{9}$. So I assume that the $0.1\times 0.2$ term above should actually be doubled. But I can't account for the factor of 2. Could someone explain this to me? Thanks!

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    $\begingroup$ Hint: you have a coin with prob head=0.3 What is the probability that, if your throw two coins, we get a tail and a head? $\endgroup$ – leonbloy Oct 19 '13 at 14:19
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Let $X_{1}$be the first tube that breaks down and $X_{2}$ the second. Then there are four possibilities:

(1) $X_{1}$ a tube of the first kind and $X_{2}$ is a tube of the first kind. Unconditional probability: $0.01$

(2) $X_{1}$ a tube of the first kind and $X_{2}$ is a tube of the second kind. Unconditional probability: $0.02$

(3) $X_{1}$ a tube of the second kind and $X_{2}$ is a tube of the first kind. Unconditional probability: $0.02$

(4) $X_{1}$ a tube of the second kind and $X_{2}$ is a tube of the second kind. Unconditional probability: $0.04$

Adding these probabilities gives $0.09$ and in the cases (1) and (4) the gadget still works. This leads to conditional probability $\frac{5}{9}$

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