Conditional Probability of decision (Sheldon Ross)

Question

I am stuck at this question from Sheldon Ross's book -

A town council of 7 members contains a steering committee of size 3. New ideas for legislation go first to the steering committee and then on to the council as a whole if at least 2 of the 3 committee members approve the legislation. Once at the full council, the legislation requires a majority vote (of at least 4) to pass. Consider a new piece of legislation, and suppose that each town council member will approve it, independently, with probability $p$. What is the probability that a given steering committee member’s vote is decisive in the sense that if that person’s vote were reversed, then the final fate of the legislation would be reversed? What is the corresponding probability for a given council member not on the steering committee?

Let C be the event that a 3 member committee person's vote is reversed and L be the event that legislation is not approved by the town council. Then, from what I understood, I need to find $P(C|L)$.

So, $P(C|L) = \frac{P(C\cap L)}{P(L)} = \frac{P(C)P(L|C)}{P(L)}$

Now, $P(L)$ is the probability that the legislation is not approved, which means it got approved by steering committee and then not approved by the council of 7. So,

$P(L) = \binom{7}{3}*P($approved by steering committee$)*P($not approved by council$)$

$ = \binom{7}{3}*(3p^2(1-p) + p^3)*\big[\binom{7}{4}(1-p)^4p^3 + \binom{7}{5}(1-p)^5p^2 + \binom{7}{6}(1-p)^6p + \binom{7}{7}(1-p)^7 \big]$

To find $P(C)$, I first need to choose the 3 members, then add the probabilities of whether 2 members approve or 3. If 2 members approve, then there are 3 ways to choose which one of the 3 members don't approve and then either the member who voted to approve may change his vote or the member who voted to not approve may change his vote. So,

$P(C) = \binom{7}{3}*3*P($either approved vote change or not approved change$) + \binom{7}{3}*P($any one of the 3 members who approved change vote$)$

$=\binom{7}{3}*3*(2p(1-p)^2 + p^3) + \binom{7}{3}*(3p^2(1-p))$

Please tell -

1. Whatever I have understood/done so far, is it correct or not?
2. How do I find $P(L|C)$?

Answer 1

I don't believe the problem is one of finding a conditional probability. It seems to me that the wording of the problem makes the reversal of some particular steering committee member's vote a counterfactual conditional that would reverse the fate of the legislation if it were to occur. That this particular person's vote is decisive in that sense is an event whose (unconditional) probability you can calculate by enumerating the outcomes for which it would occur, which is what I believe the first question is asking you to do.

Number the members of the council from $1$ to $7$ in such a way that members $1$ to $3$ are those on the steering committee. By symmetry, the probability that member $\ i$'s vote is decisive is the same for all $\ i\in\{1,2,3\}\ $, so you can answer the first question by finding the probability that member $1$'s vote is decisive. Similarly, for all $\ i\in\{4,5,6,7\}\ $, the probability that member $\ i$'s vote is decisive is the same, and the second question can be answered by finding the probability that member $4$'s vote is decisive.

Let $\ m_i=1\ $ if member $\ i\ $ approves the legislation, or $\ m_i=0\ $ if he or she doesn't, $\ a=\sum_\limits{i=1}^3m_i\ $ the number of steering committee members who approve the legislation, and $\ b=\sum_\limits{i=4}^7m_i\ $ the number of non-steering committee members who do so. The conditions for the legislation to pass are then $\ a\ge2\ $ and $\ a+b\ge4\ $.

The event that member $1$'s vote is decisive can be broken down into several cases:

• Member $1$ doesn't approve the legislation $\big($i.e. $\ m_1=0\ \big)$, $\ a=1\ $, and $\ b\ge2\ $;
• Member $1$ doesn't approve the legislation, $\ a\ge1\ $, and $\ a+b=3\ $;
• Member $1$ approves the legislation $\big($i.e. $\ m_1=1\ \big)$, $\ a=2\ $, and $\ b\ge2\ $;
• Member $1$ approves the legislation, $\ a\ge2\ $, and $\ a+b=4\ $.

In the first two cases (which are not mutually exclusive events) the legislation does not pass, but would do if member $1$ were to reverse his or her vote. In the last two cases (again not mutually exclusive) the legislation does pass, but would not do so if member $1$ were to reverse his or her vote. Thus, the probability that member $1$'s vote is decisive is \begin{align} P\big[&\big(\big\{m_1=0\big\}\cap\big\{a=1\big\}\cap\big\{b\ge2\big\}\big)\\ &\cup\big(\big\{m_1=0\big\}\cap\big\{a\ge1\big\}\cap\big\{a+b=3\big\}\big)\\ &\cup\big(\big\{m_1=1\big\}\cap\big\{a=2\big\}\cap\big\{b\ge2\big\}\big)\\ &\cup\big(\big\{m_1=1\big\}\cap\big\{a\ge2\big\}\cap\big\{a+b=4\big\}\big)\big]\\ =P\big[&\big(\big\{m_1=0\big\}\cap\big\{a=1\big\}\cap\big\{b\ge2\big\}\big)\\ &\cup\big(\big\{m_1=0\big\}\cap\big\{a=2\big\}\cap\big\{b=1\big\}\big)\\ &\cup\big(\big\{m_1=1\big\}\cap\big\{a=2\big\}\cap\big\{b\ge2\big\}\big)\\ &\cup\big(\big\{m_1=1\big\}\cap\big\{a=3\big\}\cap\big\{b=1\big\}\big)\big]\ . \end{align} The events whose disjunction appears in the probability on the right side of this identity are mutually exclusive, so the probabiliy can be evaluated by summing the individual probabilities of these four events. These are \begin{align} P\big[\big\{m_1=0\big\}\cap\big\{a=1\big\}\cap\big\{b\ge2\big\}\big]&=(1-p)^2\cdot2p\cdot\sum_{i=2}^4{4\choose i}p^i(1-p)^{4-i}\\ &=2\sum_{i=2}^4{4\choose i}p^{i+1}(1-p)^{6 -i}\\ P\big[\big\{m_1=0\big\}\cap\big\{a=2\big\}\cap\big\{b=1\big\}\big]&=(1-p)\cdot p^2\cdot 4p(1-p)^3\\ &=4p^3(1-p)^4\\ P\big[\big\{m_1=1\big\}\cap\big\{a=2\big\}\cap\big\{b\ge2\big\}\big]&=p^2\cdot2(1-p)\cdot\sum_{i=2}^4{4\choose i}p^i(1-p)^{4-i}\\ &=2\sum_{i=2}^4{4\choose i}p^{i+2}(1-p)^{5 -i}\\ P\big[\big\{m_1=1\big\}\cap\big\{a=3\big\}\cap\big\{b=1\big\}\big]&=p^3\cdot4p(1-p)^3\\ &=4p^4(1-p/)^3\ . \end{align} The event that member $1$'s vote is decisive is the sum of these four probabilities. The probability that member $4$'s vote is decisive can be found by a similar procedure. Can you try this for yourself?

Answer 2

Let the probability of approval of any legislation by a member of the steering committee/council member is $p$ and voting it out is $q= 1- p$. Then a given member of the steering committee will have a decisive vote $ E_s$ only if the other two member votes are tied (one votes for and other against). In other words, we can exclude probabilities where both these remaining members vote either for ($p^2$ ) or against the legislation ($ q^2$)

Thus, $$ P(E_s)= 1- (p^2+q^2)= 2pq= 2p(1-p) $$

Similarly, when the legislation is passed from the steering committee, then it has received either 2 votes (probability = $\binom{3}{2} p^2q= 3p^2q$ ) or 3 votes (probability = $\binom{3}{3} p^3 = p^3$). These are disjoint events. Thus, any particular council member will have a decisive vote $E_c$ when , the remaining 3 council members (not in the steering committee) vote to tie.

When, the proposal has passed with 2 votes in the steering committee $V_{2}$, then among the three council members outside steering committee, 1 must favour and other 2 not favour the proposal. When, it has passed with three votes $V_{3}$ in the steering committee, all three council members must now vote against it to tie the vote. Here we are assuming that voting preferences of the steering committee don't change during council meeting.

Thus, $$ \begin {align*} P(E_c) &= P( E_c \cap V_{2}) + P( E_c \cap V_{3})\\ &= P( E_c |V_{2})P(V_2) + P( E_c | V_{3})P(V_3)\\ &= \binom {3}{1}pq^2.3p^2q + q^3.p^3\\ &= 10p^3 (1-p)^3 \end {align*} $$ In case, the steering committee members can change their vote during council review, we can similarly work out the solution considering that 3 votes must be against and 3 in favour of a proposal before a particular member's vote becomes decisive.

$$ \begin {align*} P(E_c) &= \binom {6}{3}p^3q^3\\ &= 20 p^3 (1-p)^3 \end {align*} $$