Question

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \({{\rm{C}}_{\rm{1}}}{\rm{ = }}\){left ear tag is lost} and \({{\rm{C}}_{\rm{2}}}{\rm{ = }}\){right ear tag is lost}. Let ­ \({\rm{\pi = P(}}{{\rm{C}}_{\rm{1}}}{\rm{) = P(}}{{\rm{C}}_{\rm{2}}}{\rm{)}}\),and assume \({{\rm{C}}_{\rm{1}}}\)and \({{\rm{C}}_{\rm{2}}}\) are independent events. Derive an expression (involving p) for the probability that exactly one tag is lost, given that at most one is lost (“Ear Tag Loss in Red Foxes,” J. Wildlife Mgmt., \({\rm{1976: 164--167)}}{\rm{.}}\) (Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.)

Short answer

The expression is \({\rm{P(A}}\mid {\rm{B) = }}\frac{{{\rm{2\pi }}}}{{{\rm{1 + \pi }}}}\)

Step-by-step solution

Definition of Independent probability

If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

Drive an expression

We are given events \({{\rm{C}}_{\rm{1}}}\)and \({{\rm{C}}_{\rm{2}}}\)as well as

\({\rm{P}}\left( {{{\rm{C}}_{\rm{1}}}} \right){\rm{ = P}}\left( {{{\rm{C}}_{\rm{2}}}} \right){\rm{ = \pi }}\)

where \(\pi \ne 3.14\) (this is not the number \({\rm{\pi }}\) this is just notion, probability can not be bigger than \({\rm{1!}}\)).

The conditional probability of event A given that event \({\rm{B}}\)has occurred is the likelihood that exactly one tag is lost given that at most one is lost.

A= {exactly one tag lost};

B= {at most one tag lost}.

Given that the event \({\rm{B}}\)has occurred, what is the conditional probability of A \({\rm{P(B) > 0}}\), is

\({\rm{P(A}}\mid {\rm{B) = }}\frac{{P(A \cap B)}}{{{\rm{P(B)}}}}\)

for any two event A and \({\rm{B}}\).

Notice that event A can be written as

\(A = \left( {{C_1} \cap C_2^\prime } \right) \cup \left( {C_1^\prime \cap {C_2}} \right)\)

and the event \({\rm{B}}\), that at most one tag is lost, indicates that the left ear tag, right ear tag, or none of the tags could all be lost, equally

\(B = C_1^\prime \cup C_2^\prime = {\left( {{C_1} \cap {C_2}} \right)^\prime },\)

Apply the de morgan’s law

De Morgan's Law was invoked in this case. The likelihood we need to figure out now is

\({\rm{P(A}}\mid {\rm{B) = }}\frac{{{\rm{P(A}} \cap B)}}{{{\rm{P(B)}}}}\)

Let's calculate \(P(A \cap B)\)

\(\begin{array}{c}P(A \cap B) &=& P\left[ {\left( {\left( {{C_1} \cap C_2^\prime } \right) \cup \left( {C_1^\prime \cap {C_2}} \right)} \right) \cap {{\left( {{C_1} \cap {C_2}} \right)}^\prime }} \right]\\ &=& P\left[ {\left( {{C_1} \cap C_2^\prime } \right) \cup \left( {C_1^\prime \cap {C_2}} \right)} \right]..........(1)\end{array}\)

(1) : Because the following is true, this is correct.

\(\left( {{C_1} \cap C_2^\prime } \right) \cup \left( {C_1^\prime \cap {C_2}} \right) \subseteq {\left( {{C_1} \cap {C_2}} \right)^\prime }\)

The intersection of the two events is the smaller one, as indicated by the size of the junction. To continue from where we left off,

\(\begin{array}{c}P(A \cap B) &=& P\left( {{C_1} \cap C_2^\prime } \right) + P\left( {C_1^\prime \cap {C_2}} \right)...................(2)\\ &=& P\left( {{C_1}} \right) \cdot P\left( {C_2^\prime } \right) + P\left( {C_1^\prime } \right) \cdot P\left( {{C_2}} \right)...........(3)\end{array}\)

\(\begin{array}{c}\rm &=& \pi *(1 - \pi ) + (1 - \pi )*\pi ..........{\rm{(4)}}\\ \rm &=& 2\pi (1 - \pi )\end{array}\)

(2): We'll carry on from where we left off. The fact that the occurrences are disjointed is used here;

(3): (3): the adequate events are independent due to the independence of events \({{\rm{C}}_{\rm{1}}}\)and \({{\rm{C}}_{\rm{2}}}\) (see proposition below), and we can apply the multiplication property stated below;

(4) : we are given the probabilities in the exercise. Using that \({\rm{P}}\left( {{{\rm{C}}_{\rm{1}}}} \right){\rm{ + P}}\left( {C_1^\prime } \right){\rm{ = 1}}\)we calculate the probability that is not given (this stands for any event).

Explanation of the solution

Proposition: If \({\rm{A}}\)and \({\rm{B}}\)are independent, then

- \({A^\prime }\)and \({\rm{B}}\)are independent;

- \({A^\prime }\)and \({B^\prime }\)are independent;

- \({\rm{A}}\)and \({B^\prime }\)are independent.

Multiplication Property: Two events \({\rm{A}}\)and \({\rm{B}}\)are independent if and only if

\(P(A \cap B){\rm{ = P(A)*P(B)}}\)

We need to calculate \({\rm{P(B)}}\)as well

\(\begin{array}{c} \rm P(B) &=& P\left( {{{\left( {{C_1} \cap {C_2}} \right)}^\prime }} \right)\rm &=& 1 - P \left( {{C_1} \cap {C_2}} \right).......(1)\\ \rm &=& 1 - P \left( {{{\rm{C}}_{\rm{1}}}} \right){\rm{*P}}\left( {{{\rm{C}}_{\rm{2}}}} \right)............(2)\\ \rm &=& 1 - \pi *\pi &=& 1 - {{\rm{\pi }}^{\rm{2}}}\\ \rm &=& (1 - \pi )(1 + \pi )\end{array}\)

(1): for any event \({\rm{A,P(A) + P}}\left( {{A^\prime }} \right){\rm{ = 1}}\);

(2) : from the above-mentioned multiplication property (because \({{\rm{C}}_{\rm{1}}}\)and \({{\rm{C}}_{\rm{2}}}\)are independent).

Finally, we have

\(\begin{array}{c}{\rm{P(A}}\mid \rm B) &=& \frac{{P(A \cap B)}}{{P(B)}}\frac{{{\rm{2\pi (1 - \pi )}}}}{{{\rm{(1 - \pi )(1 + \pi )}}}}\\ \rm &=& \frac{{{\rm{2\pi }}}}{{{\rm{1 + \pi }}}}\end{array}\)

Thus, the expression is \({\rm{P(A}}\mid {\rm{B) = }}\frac{{{\rm{2\pi }}}}{{{\rm{1 + \pi }}}}\)