Let's say I will be presented with a random number from 0-9 (so 10 possibilities), but I am asked beforehand if I can make a prediction about its value.

If I think it's greater than 2 I have 70% chance of being right about that, and if I think it's less than 6 I have a 60% of being correct.

Now why aren't these chances multiplied (0.6*0.7=0.42=42%) the same as my chances of being correct when saying "between 2 and 6" (30%)?

A visual comparison of both scenarios is below:

probility overview

  • 2
    $\begingroup$ Because they are not "independent". You need to multiply the probability that it's greater than two with the probability that it's less than 6 given that it's greater than two (3/7). $\endgroup$ – David Mitra Dec 7 '11 at 19:18

Think about it this way:

What percentage of the time is it between 2 and 6? Here, we are asking for the percentage of time that it is greater than 2 and less than 6.

Now, 70% of the time it's greater than 2. Out of these outcomes, where we know it's greater than 2, only $2/7$'ths of them would be less than 6 (since there are now 7 possibilities and we want 3 of them). So, the chance of both greater than 2 and less than 6 is $(0.7)(3/7)=0.30$.

Your argument, where you multiply by .6, is incorrect because .6 is the chance that it's less than 6 in all cases. This is what I meant in my comment with "independence". You can find the probability of two events happening simultaneously by multiplying their respective probabilities if they do not influence each other's probabilities. If the two events do influence each other's probabilities (as they do here), then you have to use the so called multiplication rule: $$ P(A\text { and }B)= P(A)\cdot P(A, \text{ given that }B \text { occurred}) $$

  • $\begingroup$ Ah, so I should multiply (the chance of the number being greater than 6) by (the chance it will be less than 6 in the then remaining amount of possibilities)... Thank you very much for your explanation! $\endgroup$ – Chris Dec 7 '11 at 19:44
  • $\begingroup$ Thanks again for adding that final sentence, perfect, exactly what I was trying to describe in language (which is less clear than your formula). $\endgroup$ – Chris Dec 7 '11 at 20:01

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