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I am trying out the first probability problem at this link. I found the notations given too complicated, so I follow my own.

image

To simplify the problem's wording, the probability of the dog being alive on day n given it is alive on day n - 1 is:

P(alive, day n) = P(alive, evening n - 1)
= 1 - P(dead, evening n - 1)
= 1 - (n - 1) / (n - 1 + 2)
= [(n + 1) - (n - 1)] / (n + 1)
= 2 / (n + 1)

I got question d) right, reasoning as follow:

P(find dog, in A, day 2, alive) = P(find in A, day 2) * P(alive, day 2)
= P(find in A, day 2 | in A) * P(in A) * P(alive, day 2)
= P(find in A day 2 and not find in A day 1 | in A) * P(in A) * P(alive, day 2)
= P(find in A day 2 | in A) * P(not find in A day 1 | in A) * P(in A) * P(alive, day 2)
= 0.25 * 0.75 * 0.4 * (2/3)
= 0.05

However, when I apply similar reasoning in e), my answer do not match theirs.

P(not find dog, in A, day 2, dead) = 1 - P(find dog, in A, day 2, dead)
= 1 - P(find in A day 2 and not find in A day 1 | in A) * P(in A) * P(dead, day 2)
= 1 - (0.25 * 0.75) * (0.4) * (1 - 2/3)
= 1 - (0.25 * 0.75) * (0.4) * (1/3)
= 1 - 0.025
= 0.975

What they offer is:

P(not find dog, in A, day 2, dead) = ...
= 1 - 1/3 * [0.25 * 0.333]
= 35/36
= 0.9722

This is where my answer differs from theirs:

0.4 * 0.75 vs 0.333

I do not understand where they got that 0.333 from. For my answer, P(in A) = 0.4, P(not find | looked in A) = 0.75. This is the same reasoning I use in question d) which gives an answer that agrees with theirs. Can someone please help me? TIA!

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    $\begingroup$ The problem with using your own notation is that the only person who wants to use it is you. People are too busy to translate your work and will simply pass it by. $\endgroup$ – Paul Feb 1 at 9:55
  • $\begingroup$ My notations are way simpler, and that should help busy people. $\endgroup$ – muxo Feb 1 at 10:03
  • $\begingroup$ You're wasting our time by posting two questions with the same mistakes without linking them: math.stackexchange.com/questions/3530242. $\endgroup$ – joriki Feb 1 at 11:27
  • $\begingroup$ @joriki, I am not wasting your time. I split it out, asking about e) in this post. The other question is for d). $\endgroup$ – muxo Feb 1 at 11:30
  • $\begingroup$ You're indeed wasting our time. The posts contain the same wrong calculation at the beginning. I took time not only to point out that mistake to you under the other post but also to explain again when you insisted there wasn't one. People here can't profit from that effort when you don't link the two posts that share that content. $\endgroup$ – joriki Feb 1 at 11:32
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I checked only (e), after I did my own calculation and got the same result as in the pdf.

You caclulated

P(find dog, in A, day 2, dead)

correctly as $0.025$. And then used $1-0.025$ to calculate the probability that he did not find the the dead dog. But you forgot to include the condition that you already know that Oscar didn't find his dog on the first day.

So if we make the events

X: "Oscar did find his dead dog, when searching A on second day"

and

Y: "Oscar did not find his dog, when searching A on first day"

then you calculted $P(X)=0.025$ correctly, but the task was to find the conditional probability $P(X|Y)$. The formula for that is

$$P(X|Y) = \frac{P(X \cap Y)}{P(Y)}.$$

Now in our case we have $X \subseteq Y$. If Oscar found his dead dog on day 2, he couldn't have found it on day 1. So the above simplifies to

$$P(X|Y) = \frac{P(X)}{P(Y)} = \frac{0.025}{P(Y)}.$$

So what is $P(Y)$? Well, if the dog is actually in B, then Oscar will not find in A with probability $1$. And if the dog is in A, still the chance to not find it on a given day is $0.75$. Since the dog being in A and B is mutually exclusive and nothing else is possible, we get

$$P(Y)=0.6\times1+0.4\times0.75=0.9$$

So the probability $P(X|Y)$ is $\frac{0.025}{0.9}=0.02777\ldots$ Which leads directly to the result in the pdf $0.97222\ldots$.

That factor of $0.9$ is exatly what you also found missing:

$$\frac{0.4\times0.75}{\frac13}=0.9$$

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  • $\begingroup$ Succinct! Much sense! Such unequivocal! Wow! $\endgroup$ – muxo Feb 1 at 12:09

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