Long Division

The first step is to find out the probabilities that the member is a Conservative of Labour member. The odds of a Conservative member voting the same way twice is $1/1$ , whereas the odds of a Labour member voting the same way twice is $1/2$ .

The unnormalized probability of this member being a Conservative member is $1/1 * 1/3$ , and the unnormalized probability of this member being a Labour member is $1/2 * 2/3$ . Both of these are equal to $1/3$ , so the normalized odds for the member being a Conservative or Labour member is $1/2$ and $1/2$ respectively.

The odds that a Conservative member will vote the same way next time is $1$ , and the odds that a Labour member will vote the same way is $1/2$ . Therefore, the final odds of this member voting the same way is $1*1/2 + 1/2*1/2 = \boxed{0.75}$ .

Let $C$ denote the event that a random person chosen from the parliament represents a conservative member and let $L$ denote the event that the person is a labour member.

Now, $P(C) = \dfrac{1}{3}$ and $P(L) = \dfrac{2}{3}$

Let $SS$ be the event that two successive votes have been voted for. We want to find the probability that the person will vote the same way the next time. (The $S$ denoting successful vote)

There are possibilities for who the person is - Conservation or Labour member. We will find the likelihood of the random person being either of those and then we can predict what the likelihood of no change will be in the next session of voting.

Thus, we wish to find

$P( C | SS)$ and $P ( L | SS)$ . We will find just $P( C | SS )$ and then $P( L | SS) = 1 - P( C | SS)$

Thus, by the bayes rule,

$P( C | SS ) = \dfrac{P(SS | C) \times P(C)}{P(SS)} = \dfrac{1 \times \dfrac{1}{3}}{1 \times \dfrac{1}{3} + \dfrac{1}{2} \times \dfrac{1}{3}} = \dfrac{1}{2}$

Thus, $P ( L | SS) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$

Now, the probability that the next vote will be the same as the previous one is:

• If the person is a conservative, it will be $1$ since he never changes his mind.
• If the person is a Labour member, it will be $\dfrac{1}{2}$ , since he randomly changes his mind.

Now we have already calculated the probabilities of both these situations occurring, and thus we get

$P ( L | SS) \times P(\text{the labour member will change his mind}) + P ( C | SS) \times P(\text{the conservative member will change his mind})$

Plugging in the values we get,

$\dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{1}{2} \times 1 = \dfrac{1}{4} + \dfrac{1}{2} = \boxed{0.75}$

Denote the event that the chosen member is Conservative with $C$ , and the event that the chosen member votes the same twice with $S$ .

To solve the problem, we will first find the probability that the chosen member is Conservative based on the fact that he voted the same twice (ie. $P(C|S)$ ). Using the definition of conditional probability, we have that,

$P(C|S) = \frac{P(C \cap S)}{P(S)}$

The probability that the chosen member will be Conservative and that he will vote twice the same is simply $^1\!/_3$ , from what we were given. Now to find $P(S)$ , we will consider whether the member is Conservative or Labour: $P(S) = P(C) \cdot 1 + P(\neg C) \cdot \frac{1}{2} = \frac{1}{2}$ . We can now return to $P(C|S)$ .

$P(C|S) = \frac{\:^1\!/_3\:}{\:^2\!/_3\:} = \frac{1}{2}$

So the probability that the chosen member will vote the same a third time, considering whether the chosen member is Conservative or not, is $\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} = \boxed{0.75}$

Suppose there are P members of parliament. We know that P/3 members are Conservative and would be observed to vote the same way twice (Conservatives will always vote the same way twice), and P/3 members are Labour members and would vote the same way twice (Labour members have a 50% chance of voting the same way twice so you would expect that half of them would do so). *Hence the randomly chosen member has an equal chance of being either Labour or Conservative. *

The probability that the member votes the same way next time is:

The probability that the member is Conservative and will vote the same way + the probability that the member is a Labour member and will vote the same way again

Which is equal to (1/2)(1) + (1/2)(1/2) = 3/4 = 0.75