0
$\begingroup$

I've got a question as follows: A heavy container with chemical waste has been illegally dumped into the sea, and is now lying in a fixed but unknown position on the bottom of the sea. The container is believed to be situated in British territorial waters with probability 0.8. The Royal Navy are planning a search through the British territorial waters only, to try to find the container. If the container is situated in British territorial waters, then the probability that it will be found by such a search is 0.7 (while the probability that it will be overlooked is 0.3).

(i) Calculate the probability that the container will be discovered by this search through the British territorial waters

(ii) Suppose that this search through the British territorial waters turns out to be unsuccessful, i.e. the container is not found. Given this information, what is now the probability that the container is situated in British territorial waters?

...........

For part (i) I think you use the conditional probability formula (P A given B) = P(A intersection B) / P (B). In this context, P(Found given British) = 0.7 and P(B) = 0.8 hence the answer is 0.8 x 0.7 = 0.56. Can someone confirm this?

For part (ii) I'm quite clueless about what to do

Thanks in advance

$\endgroup$
0
$\begingroup$

I drew a tree diagram concerning this question to help visualize.

Part I.

Yeah, that looks good.

Part II.

In essence, what this question is asking is "GIVEN that the container is not found, what is the probability that it is in British waters?"

The second and fourth branches on the tree diagram are $F'$, which means it isn't found. This means the overall possibility that the container is not found is $0.8 \times 0.3 + 0.2 \times 0.3 = 0.3$. Then, the other part of the question asks for the probability that it is in British waters - that's the second branch only. Now we can define these:

$P(A) =$ Not found $= 0.3$

$P(B) =$ In British waters $=0.8$

$P(A \cap B) =$ Not found and in British Waters $= 0.24$

Thus:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ $$P(A|B) = \frac{P(0.24)}{P(0.8)}$$ $$P(A|B) = 0.192$$

$\endgroup$
1
  • $\begingroup$ You have defined P(A) as the probability the waste is not found, but you have not conditioned on the location of the waste being in the British Territorial Waters. This propagates an error through the rest of your calculation. P(A) is subject to the Law of Total Probability $\endgroup$
    – DWD
    Apr 14 '17 at 12:50
0
$\begingroup$

Denote:

  • Event B as the event the Waste is in British waters
  • Event F as the search finds the waste
  • $P(B)=0.8$,$P(B^C)=0.2$,
  • $P(F)=0.7$,$P(F^C)=0.3$

Part one:

For part (i) I think you use the conditional probability formula (P A given B) = P(A intersection B) / P (B). In this context, P(Found given British) = 0.7 and P(B) = 0.8 hence the answer is 0.8 x 0.7 = 0.56. Can someone confirm this?

Yes: expressing the probability of an event IF some other event does or does not occur, is the application of conditioning on information. Usually we use the word "given". "A given B occurs". It is the same as "A if B occurs". $$P(F | B)= \frac{P(F \cap B)}{P(B)}=0.7$$

$$P(F \cap B) = P(F | B)P(B)=0.7 \times 0.8$$

Part two: $$P(F^C|B)=0.3$$ Find $P(B|F^C)$ $$P(B|F^C)=\frac{P(B\cap F^C)}{F^C}=\frac{P(F^C|B)P(B)}{F^C}=\frac{P(F^C|B)P(B)}{P(F^C|B)P(B)+P(F^C|B^C)P(B^C)} $$ The denominator in the last fraction makes use of the Law Of Total Probability. $$P(B|F^C)=\frac{0.3\times 0.8}{0.3\times 0.8+1\times0.2}=\frac{6}{11}$$

Hope you find this useful

$\endgroup$
0
$\begingroup$

For Your Part two I would recommend Baye's Theorem.

$P(A)$= Chemical in British Water (0.8)
$P(E)$=Search unsuccessful (0.3) $$P(A|E)=\frac{P(A).P(E|A)}{P(A).P(E|A)+P(A^{'}).P(E|A^{'})}$$

$$P(A|E)=\frac{0.8*0.3}{0.8*0.3+0.2*1}$$ $$P(A|E)=\frac{6}{11}$$

$P(E|A^{'})=1$ Because Vessel can't search for a vessel which ain't there.
PS: Your (i) looks good enough
EDIT: Thanx for pointing out the calc. error.

$\endgroup$
1
  • 1
    $\begingroup$ I believe your final value for P(A|E) should be 6/11 $\endgroup$
    – DWD
    Apr 14 '17 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy