If that happens then what denies that this happens? The conclusion holds in exac

Question

If that happens then what denies that this happens? The conclusion holds in exactly the same three possibi lities in which the premise holds, and so the two assertions They mistakenly envisage, not the denial of the con should have the same probability. Suppose the two coins ditional, but the denial of its consequent given are tossed fairly. The first assertion holds in three out of cedent 19) its ante- the four equally likely outcomes, and so it has a probability of 8/4. But, the probability that the Problem 7: what is the probability of a conditional? dime came down heads given that the nickel didn't come down heads is only 1/2.Something has gone wrong. Andthe The principle to which we have just alluded also explains only viable diagnosis is that it is the answer that individuals give to the following sort of erroneous to assume that the the conditional. conditional probability corresponds question: to the probability What is the probability that ifthe nickel isheads then No one doubts that some people some of the time make the dime is heads? other estimates of the probability of a conditional (29,32- They interpret the question to mean 34]. The model theory offers an explanation. As in other forms of reasoning, different individuals adopt different If the nickel is heads then what's the probability that strategies One strategy contrasts the the dime is heads? cit mental model of a conditional with the alternative in which the conditional is false (Box 2.This 'equiprobable' They estimate, not the probability ofthe conditional, but strategy is analogous to the suppositional theory B7,381. It the conditional probability of its consequent given its too yields a probability of 12. Asecond strategy treats the antecedent 19l. Formal rule theories do not seem to have conditional as though it were a conjunction, again reflect- addressed this question, but several probabilistic theories ing its single explicit mental model. It yields a probability of reasoning have postulated that the probability of a of 1/4. And a third strategy sums the probabilities of all three possibilities in which the conditional holds. It yields conditional is, and should be, the aforesaid conditional probability 130,31]. It is also a consequence of the supposi- probability of 34, which matches the required probability of the premise in the inference above. tional theory 16,71 No-one doubts that some people some of the time make this estimate [29,32-34]. But, is it correct? Here is one Conclusions argument to the contrary, which hinges on this inference We have described seven problems about conditionals that concern their meanings, their truth or falsity, their coun- The nickel came down heads or the dime came down terfactual use, their denials, their probabilities and the heads, or both did factors that affect reasoning from them. They are not the only problems, ofoourse (Box4), but any adequate theory of Therefore, if the nickel didn't come down heads then conditionals has to solve them. Table 2 summarizes the solutions of the three main theories. the dime did.

Explanation / Answer

Firstly, I will breif you what is conditional probability and give an example to you

Suppose there are two events p(A) and p(B), we phrase a probability case in such a way that when probability of occurence of p(B) provided p(A) occurs.

The above statements stresses on the occurences of the p(A) first. If p(A) did'nt do, there is no need of calculating p(B).

If p(A) does, then we calculate p(A)intersection p(B), i.e. intersection gives the relation that occurence of both p(A) and p(B) together

Hence we conclude that

p(B/A) = p (B intersection A)/ p(A)

eg. Consider you are tossing two coins, p(Heads) = 0.5, p(Tails) = 0.5

I want to calculate the probability that both the coins shows different ones provided it must be atleast thrown with one of the heads

Here there are two events

p(B) = that both the coins shows different ones

p(A) = atleast thrown with one of the heads

p(A) = 3/4

p(B) = 1/2

also p(A intersection B) = 1/2

hence p(B/A)

= p(B intersection A) / p(A)

= 0.5/0.75 = 0.666

Also, in the last paragraph it mentions about the strategies

--> equiprobable startegy

--> treating conditional as a conjuction (and) , refelcting its single explicit mentalmodel

--> Sums the probaility of all the three possibilities in which conditional holds

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