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I am trying to solve the following problem.

A fisherman is equally likely to go fishing at one of the three ponds $A,B,C$. The probability to catch fish if he cast his rod at pond $A$ is $0.4$, at pond $B$ it is $0.5$ and at pond $C$ it is $0.3$. It is known that the fisherman cast his rod three times and caught only one fish. Find the probability that he went fishing at pond $A$.

The answer given is $0.346$.

Here is my thinking.

Let $P(A) = P(B) = P(C) = \frac{1}{3}$ be the probability to go fishing at $A$, $B$ or $C$ respectively.

Let $P(F|A) = 0.4$, $P(F|B) = 0.5$, and $P(F|C) = 0.3 $ be the conditional probabilities to catch fish if he went to $A$,$B$ and $C$ respectively.

Using all that information, I could use the total probability formula to compute the conditional probability that he went to pond $A$ given he caught fish $$ P(A|F) = \frac{P(F|A) P(A)}{P(F)} \\ P(F) = P(A) P(F|A) + P(B) P(F|B) + P(C) P(F|C) $$

The same is true for $P(B|F)$ and $P(C|F)$. Unfortunately this is not what is needed. I don't know how the fact that he cast his rod three times has to be taken into account. I would appreciate your help.

SOLUTION: Here is the solution, based on the original idea to use the total probability and the idea coming form the comments to consider the probability of going to pond $X$ and catching 1 fish in 3 attempts $P(1:3|X)$. $$ P(1:3|X)=\frac{3!}{2!1!}S*(1-S)^2 $$ Here $S$ are the probabilities $P(F|A) = 0.4$, $P(F|B) = 0.5$, and $P(F|C) = 0.3 $. Therefore, I should use $$ P(A|1:3)=\frac{P(1:3|A)*P(A)}{P(1:3)} $$ Where $$ P(1:3)=\frac{3!}{2!1!}\left(P(A)P(F|A)(1-P(F|A))^2 + P(B)(1-P(F|B))^2 + P(C)(1-P(F|C))^2\right) $$ Substituting this will give the final answer.

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  • $\begingroup$ Have you considered making a tree diagram? Alternatively, instead of considering $P(F|A)=0.4$, consider $P(\text{"caught one fish in three attempts"}|A)$ and calculate that. $\endgroup$ – JMoravitz Aug 28 '15 at 17:17
  • $\begingroup$ I am afraid that I don't know how to calculate this. Perhaps this is what is missing. I am looking for three diagrams online. $\endgroup$ – Alexander Cska Aug 28 '15 at 17:19
  • $\begingroup$ Hint: One throws a coin three times with probability h for head and t=1-h for tails, what is the probability to obtain one head and two tails? $\endgroup$ – Did Aug 28 '15 at 17:22
  • $\begingroup$ Shouldn't that be $P(h)*P(t)*P(t)=3*h*(1-h)^2$? $\endgroup$ – Alexander Cska Aug 28 '15 at 17:31
  • $\begingroup$ ?? No this is not P(h)P(t)P(t), yes this is 3h(1-h)^2. (And please use @.) $\endgroup$ – Did Sep 1 '15 at 15:14
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The probability of catching exactly one fish are shaded below in Venn diagrams for each pond separately, where the three circles for a given pond correspond to catching a fish on the first, second, and third tries, respectively. (Areas not to scale.)

For Pond B, where the calculation is partly shown, the probability of catching exactly one fish is $3\cdot\left(.5-2\times.5^2+.5^3\right)=.125$, and similarly, for Pond A, it’s $3\cdot\left(.4-2\times.4^2+.4^3\right)=.144$, and for Pond C, it’s $3\cdot\left(.3-2\times.3^2+.3^3\right)=.147$.

The probability you seek is the probability that you’re in the green area, given that you’re in the shaded area. Since you’re equally likely to be in any of the three ponds, the areas need not be weighted, and the probability is $\frac{.144}{.125+.144+.147}=.34615\dots$

enter image description here

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One interpretation of the problem is that every time she casts her rod say at Pond A, she has probability $0.4$ of catching a fish. We will assume, unreasonably, that the results of the casts are independent. Then given that she cast $3$ times in Pond A, the probability she caught exactly $1$ fish is $\binom{3}{1}(0.4)^1(0.6)^2$.

Compute similar probabilities for the other two ponds, and then use the same method as the one you used, but with the conditional probabilities of catching exactly one fish.

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There is insufficient information to answer the question.

What is missing is the relationship between the statement "the probability of catching a fish" and "the fisherman cast his rod three times and caught only one fish." Are we to assume, for example, that the probability of $0.4$ for pond $A$ is the per-casting probability, or the probability of catching ANY number of fish (i.e., at least one fish) once the fisherman chooses that pond? Either interpretation will give different answers. Without clarifying what the meaning of these probabilities are, it is not possible to proceed further.

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  • $\begingroup$ If he casts his rod , say at A, with probability 0.4 he will catch a fish. I edited the question. $\endgroup$ – Alexander Cska Aug 28 '15 at 17:34
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You can supply some of the missing information through plausible assumptions explicitly stated and the get answers conditioning on those assumptions that will apply or not depending on the verifiability of those assumptions.

For instance, you can explicit the assumption that, given the chosen lake, each cast will be a realization of an independent Bernouli with probability given by the chosen lake.

So you can (verbose mode activated for clarity's sake):

  1. define $L$ as $$ L = \begin{cases} 1, &\text{if $A$ occurs} \\ 2, &\text{if $B$ occurs} \\ 3, &\text{if $C$ occurs} \end{cases} $$ with $p_1 := P(L = 1) = P(A)$, $p_2 := P(L = 2) = P(B)$, and $p_3 := P(L = 3) = P(C)$;

  2. define $X_i$ as an indicator of catching fish in the $i$-th attempt, $i = 1, 2, 3$, with $X_1$, $X_2$ and $X_3$ independent and identically distributed given $L$ so that, for $i = 1, 2, 3$, \begin{align*} q_1 := P(X_i = 1|L = 1) & = 1 - P(X_i = 0|L = 1) = P(F|A) \\ q_2 := P(X_i = 1|L = 2) & = 1 - P(X_i = 0|L = 2) = P(F|B) \\ q_3 := P(X_i = 1|L = 3) & = 1 - P(X_i = 0|L = 3) = P(F|C); \end{align*}

  3. define $X = X_1 + X_2 + X_3$, so that $(X|L=\ell) \sim \text{Bin}(3,q_\ell)$ and $P(X=x|L=\ell) = \binom{3}{x} q_\ell^x (1 - q_\ell)^{3-x}$

Now, you can formulate your question properly without getting tangled with informal language.

Let's see: with those terms, you asked for $P(L = 1|X = 1)$, which you can get now with an application of good ol' Bayes' Theorem \begin{align*} P(L = 1|X = 1) & = \frac{P(X = 1|L = 1) P(L = 1)}{P(X = 1)} \\ & = \frac{P(X = 1|L = 1) P(L = 1)} {P(X = 1|L = 1) P(L = 1) + P(X = 1|L = 2) P(L = 2) + P(X = 1|L = 3) P(L = 3)} \\ & = \frac{\binom{3}{1} q_1 ^ 1 (1 - q_1) ^ {3 - 1} p_1} {\binom{3}{1} q_1 ^ 1 (1 - q_1) ^ {3 - 1} p_1 + \binom{3}{1} q_2 ^ 1 (1 - q_2) ^ {3 - 1} p_2 + \binom{3}{1} q_3 ^ 1 (1 - q_3) ^ {3 - 1} p_3} \\ & = \frac{q_1 (1 - q_1) ^ 2 p_1} {q_1 (1 - q_1) ^ 2 p_1 + q_2 (1 - q_2) ^ 2 p_2 + q_3 (1 - q_3) ^ 2 p_3} \\ & = \frac{q_1 (1 - q_1) ^ 2 1/3} {q_1 (1 - q_1) ^ 2 1/3 + q_2 (1 - q_2) ^ 2 1/3 + q_3 (1 - q_3) ^ 2 1/3} \\ & = \frac{q_1 (1 - q_1)^2} {q_1 (1 - q_1)^2 + q_2 (1 - q_2)^2 + q_3 (1 - q_3)^2} \\ & = \frac{0.4 \times 0.6 ^ 2} {0.4 \times 0.6 ^ 2 + 0.5 \times 0.5 ^ 2 + 0.3 \times 0.7 ^ 2} \\ & = 0.3461538 \end{align*} as stated initially.

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