# conditional probability example from tsitsiklis probability book

I am working on tsitiklis probability book. and there is this solved example in the book which i cannot understand. Please

A conservative design team, call it C, and an innovative design team, call it N, are asked to separately design a new product within a month. From past experience we know that: (a) The probability that team C is successful is 2/3. (b) The probability that team N is successful is 1/2. (c) The probability that at least one team is successful is 3/4. Assuming that exactly one successful design is produced, what is the probability that it was designed by team N?

Then it goes one and states:

There are four possible outcomes here, corresponding to the four combinations of success and failure of the two teams: SS: both succeed, FF: both fail, SF: C succeeds, N fails, FS: C fails, N succeeds.

It further states:

We were given that the probabilities of these outcomes satisfy:

P(SS) + P(SF) = 2/3 , P(SS) + P(FS) =1/2. P(SS) + P(SF) + P(FS) =3/4.

and from these relations, together with the normalization equation P(SS) + P(SF) + P(FS) + P(FF) = 1

we can obtain the probabilities of individual outcomes

P(SS)= 5/12, P(SF)=1/4, P(FS)=1/12, P(FF)=1/4.

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my question is how does he do this ? and i don't get it that

P(SS) + P(SF) + P(FS) =3/4. is given ? how does he compute this ?

From the question, we can only infer that P(SF) = 2/3 and that P(FS) =1/2.

I don't get it how he infers that P(SS) + P(SF) = 2/3

Any advise ? And then i just don't get it how he computes

P(FS) = 1/12 etc...

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He stated at least one team successful has probability 3/4. Isn't that just P(SS)+P(SF)+P(FS) ? – Raskolnikov Oct 27 '13 at 6:59
actually in his proposed solution he gives all three eqns ... additionally, i completely agree that the only bit of information matching a correct equation is P(SS)+P(SF)+P(FS)=3/4. how does he get the other ones ? – Richard Oct 27 '13 at 8:08

## 1 Answer

You have "(a) The probability that team C is successful is 2/3." This does not tell you anything about Team I.

So you might write it as something like $\Pr(S?) = \frac23$ or as $\Pr(SS) + \Pr(SF) = \frac23$.

You also have "(c) The probability that at least one team is successful is 3/4."

You might write this as something like $\Pr(S? \text{ or }?S) = \frac34$ or $\Pr(SS) + \Pr(SF) + \Pr(FS) = \frac34$.

Combining these two gives $\Pr(FS) = \frac34- \frac23 = \frac{1}{12}$ and you can find the others in a similar way, as you also have $\Pr(SS) + \Pr(SF) + \Pr(FS) + \Pr(FF) = 1$ and so four simultaneous equations in four unknowns.

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rite is rong and I do not know what P(s) stands for either. The probability that Team C is successful whether or not Team N is successful is $\frac23$ and I have written this as $\Pr(SS)+\Pr(SF)=\Pr(S?)$. Similarly the probability that Team N is successful whether or not Team C is successful is $\frac12$ and I have written this as $\Pr(SS)+\Pr(FS)=\Pr(?S)$. There is no independence here. – Henry Oct 28 '13 at 1:30
but these events (SS) and SF are mutually exclusive ? obviously. I thought independence kicks in since SS means Team C successful intersection Team N successful. Is this incorrect ? So, P(S_c S_n) = P(s_c) \times p(s_n). P(S_c) = probability that team C is successful. p(s_n) probability that team N is successful. – Richard Oct 28 '13 at 2:35
Your changes to notation are slightly confusing, but independence would require $\Pr(S_c \text{ and } S_n) = \Pr(S_c ) \times \Pr(S_n)$. That is not true here since $\frac{5}{12} \not = \frac{2}{3} \times \frac{1}{2}$ – Henry Oct 28 '13 at 12:25