QUESTION 3 Where, in these 6 panels, does the fathers biggest misconception lie?

Question

QUESTION 3 Where, in these 6 panels, does the fathers biggest misconception lie? THERE ARE THREE CANDIDATE IS GOING TOSTATES VOTING TODAY OGE TODKYS PRIMARIG6HER OPPONENT HAS A GO% CHANCE TO WIN EACH OF THEM NONSENSE UT, THAT MEANSSINCE PROBABIUTIES HAVE TO CHANCSOF SUN TO 100%, AND THERE ARE NNG Au ONLY TWO PEOPLE RUNNING MY CANDIDATE HAS A 724% THREE IS CO% OF 6 a1.6%, THREE STATES HBOMA4MOM, YOU WERE AT LEAST WHEN ALL WRONG ABOUTI GO To KIL HIM Information in Panel 2 is invalid. Reasoning in Panel 3 is flawed. Reasoning in Panel 4 is flawed. The father is right, there is no error.

Explanation / Answer

The father's biggest misconception lies in 4th panel. Thus, 3rd option is correct: "Reasoning in Panel 4 is flawed".

Reason: The opponent has 60% chance or probaility of winning in each state. It means father's candidate has only 40% to win in each state because the total probability is one. Now, the probability of winning by father's candidate in all the three states is 40%*40%*40% = 6.4% whereas that of opponent is 21.6% as given in panel 3 and it is correct.

But now the sum of 6.4% and 21.6% is only 28%, that is, 0.28 but not 1. It need not be one because, in probability, the sum of probabilities of all exhaustive events must be one. And here, the events of each candidate winning all three states are not exhaustive events. The exhaustive events means all possible events or outcomes and when you add the probabilities of all these outcomes, it must be one.

Now, let us consider all exhaustive events which are as follows:

1. winning by father's candidate in 0 states, that is, not winning in any state.

2. winning by father's candidate in 1 state

3. winning by father's candidate in 2 states.

4. winning by father's candidate in all three states.

These 4 are the exhaustive events. Now, there are only two outcomes, winning or not winning with probablities 0.4 and 0.6 respectively for the father's candidate.Thus, it follows binomial distribution with n = 3, that is, total states.

Formula: P(X=r) = nCr pr qn-r

where P(X=r) is the probability of 'r' successes, ,i.e., winning in 'r' states by father's candidate, p is the probability of winning in single state = 0.4 and q is the probability of not winning in single state = 0.6

Let us add all the probabilities.

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 3C0 0.40 0.63 + 3C1 0.41 0.62 + 3C2 0.42 0.61 + 3C3 0.43 0.60 = 0.216+0.432+0.288+0.064 = 1.

Similarly, if you add all the probabilities of opponent candidate, it will be 1. Infact, if father's candidate is not winning in all three states implies that the opponent is winning in all three states, if father's candidate is winning in one state implies that the opponent is winning in other two states, and the like. So, it is already calculated above and the only thing we have to change is that we need to write P(X=3) first and proceed backwards and end with P(X=0) last.And the calculation remains same for opponent.and it is one. So, we need not calculate for both candidates, when we calculate for one candidate the other candidate's probabilities are automatically included and thus all exhaustive events are considered.

Father's candidate:    Opponent:

P(X=0) = P(X=3)

  P(X=1) =  P(X=2)

P(X=2) =  P(X=1)

   P(X=3) =  P(X=0)

Note: If you do for opponent, then p=0.6 and q=0.4.

So, finally, the point which the father missed is that he considered outcomes of single state and three states are same but it is not same. Because, in one state, there are only two possibilities for a candidate winning in that one state or not winning in that one state but when states are increased  to 3, the possibilities  increased to 4 as mentioned above.

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