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.
=—————-
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…
Best 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.