Question

A family dog is missing after a picnic in the park. Three hypotheses are suggested: A : it has gone home, B: it is still enjoying the big bone in the picnic area, C: it has gone into the woods in the park. The a priori probabilities of the above hypotheses, which are assessed from the habits of the dog, are estimated to be respectively. Two children are sent to look for the dog. The first child returns to the park to search the picnic area and the woods. If the dog is in the picnic area, there is 80% chance that it will be found. The chance drops down to 40% if the dog has gone into the woods. The other child goes back home to have a look. (i) What is the probability that the dog will be found in the park? (ii) What is the probability that the dog will be found at home? (iii) Given that the dog is found in the park, what is the probability that it is indeed found in the picnic area? (iv) Given that the dog is lost, what is the probability that it is lost in the woods?

Answer 1

Answer:

(i) $\frac {4}{15}$

(ii) $\frac 1 3$

(iii) $\frac 2 3$

(iv) $\frac 3 4$

Step-by-step explanation:

Let $A$ be the event that the dog has gone home,

$B$ be the event that the dog has gone to the picnic area, and

$C$ be the event that the dog has gone to the park.

Assuming all the three events are equally likely to happen, so,

$P(A)=P(B)=P(C)=\frac 1 3\;\cdots (1)$

Now, let $F$ and $L$ bet the event of found and lost of the dog.

Given that the chances of finding the dos are $80\%$ and $40\%$, if the dog is in the picnic area and woods respectively.i.e

$P\left(\frac {F}{B}\right)=0.8\;\cdots (2)$,

$P\left(\frac {L}{B}\right)=1-0.8=0.2\;\cdots (3)$ and

$P\left(\frac {F}{C}\right)=0.4\;\cdots (4)$

$P\left(\frac {F}{C}\right)=1-0.4=0.6\;\cdots (5)$

(i) The probability that the dog will be found in the park

$=$ Probability of going the dog to park $\times$ Probability of found in the park

$=P(B)\timesP\left(\frac {F}{B}\right)$ [using equations (1) and (2)]

$=\frac 1 3 \times 0.8$

$=\frac {4}{15}$

(ii) If the dog is in home, the chance of finding the dog is 100%.

So, the probability that the dog will be found at home

$=$ The probability that the dog has gone home

$=P(A)$

$=\frac 1 3$ [ from equation (1)]

(iii) Given that the dog is found in the park, so, the probability of founding the dog in the picnic area of the park

$=\frac {P(B)\timesP\left(\frac {F}{B}\right)}{P(B)\timesP\left(\frac {F}{B}\right)+P(C)\timesP\left(\frac {F}{C}\right)}$

$=\frac {\frac 1 3\times0.8}{\frac 1 3\times0.8+\frac 1 3\times0.4}$ [using equations (1), (2) and (4)]

$=\frac 2 3$

(iv) Given that the dog is lost, so, the probability of losing the dog in the woods

$=\frac {P(C)\timesP\left(\frac {L}{C}\right)}{P(B)\timesP\left(\frac {L}{B}\right)+P(C)\timesP\left(\frac {L}{C}\right)+P(A)\timesP\left(\frac {L}{A}\right)}$

$=\frac {\frac 1 3\times0.6}{\frac 1 3\times0.2+\frac 1 3\times0.6+\frac 1 3 \times 0}$ [using equations (1), (3) and (5)]

$=\frac 3 4$ .