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$X=X_1+X_2+X_3$ , where $X$ is the expectation for the three triplets together , and $X_i$ is the expectation for the triplet $X_i$ ,

$EX_1=EX_2=EX_3=\frac{\binom{2}{1}\binom{4}{1}\binom{3}{1}}{\binom{9}{3}}=\frac{24}{84}=\frac{2}{7}$

$EX=EX_1+EX_2+EX_3=3*\frac{2}{7}=\frac{6}{7}$

No. of selecting 3 members from 9 is 9C3 =84

Possibilities in which there is one parent, one son and one daughter is 2×3×4 =24 as there are 2 parents 3 sons and 4 daughters.

As there are 3 triplets, any of them can have this unique triplet.

Therefore probability is (24×3)/84 = 6/7

It is right that the probability for a certain triplet to be made up of {p,d,s} is $\frac{24}{84}=\frac{2}{7}$ . But since the probabilities for different triplets to be a psd-triplet are dependent, we cannot just say the answer is $3×\frac{2}{7}$ .

The expectancy value for a quantity q is defined as $Q_{expected}=\sum_i {Q(A_i) P(A_i) }$ , where the $A_i$ are the eventualities, $Q$ their associated quantities and P their probabilities. In our case we let the eventualities $A_i$ be defined by the number of psd-triplets.

We list the possible ways:

The trouble is that $P_1$ and $P_2$ are unknown, and not very easy to calculate. What we do know however is the eventualities where a certain triplet is a psd-triplet is $P_1+2P_2+P_3=\frac{2}{7}$ and, most importantly, that $P_3=0$ since there are only two parents.

Only thanks to the fact that $P_3=0$ , and using $P(1)=3P_1$ and $P(2)=3P_2$ we can say that $Q_{expected} = ((0 P(0)+1 P(1)+ 2P(2)+3P(3)= 3P_1+ 6P_2+3P_3=3(P_1+2P_2+P_3)=3×\frac{2}{7}=\boxed{\frac{6}{7}}$ .