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2. Hikers in remote regions sometimes carry with them an emergency locater. Once activated, a locater emits signals that help search parties in case of an emergency. The following table gives the probabilities of finding lost hikers within the first 24 hours of activation. Locater No Locater Row Total 0.10 Found Not Found 0.04 0.16 Column Total 0.7426 Grand Total -1 0.80 0.20 0.70 (a) Find the probability of a lost hiker wearing a locater and not being found. Answer: 0.04 (b) From a total of 1,000 lost hikers, how many would be expected to wear a locater? Answer: 740 (c) From a total of 1,000 lost hikers who don't wear a locater, how many are expected to be found? Answer: about 385 (d) If a lost hiker is found, what is the probability that the hiker was wearing a locater? Answer: 7/8 (e) Produce a tree diagram for the information in the table. Write probabilities on the branches for each of the stages. (3.5.4)

Answer 1

Let

• L be the event that a lost hiker wears an emergency locator.
• NL be the event that a lost hiker does not wear an emergency locator.
• F be the event that a lost hiker is lost
• NF be the event that a lost hiker is not found

From the table we know the following joint probabilities (cell values of the table)

$\begin{align*} P(L\cap F)=0.70\\ P(L\cap NF)=0.04\\ P(NL\cap F)=0.10\\ P(NL\cap NF)=0.16\\ \end{align*}$

We also know the following marginal probabilities (column and row totals)

$\begin{align*} P(L)=0.74\\ P(NL)=0.26\\ P(F)=0.80\\ P(NF)=0.20 \end{align*}$

a) The probability of a lost hiker wearing a locator (event L) and not being found (event NF) is

$\begin{align*} P(L\cap NF)=0.04\quad \text{value under column locator and row "not Found"} \end{align*}$

b) the probability that a lost hiker would wear a locator (event L) is the marginal probability of L

$\begin{align*} P(L)=0.74\quad \text{value under column total for locator} \end{align*}$

From a total of 1000 lost hikers, the number expected to wear a locator is

$\begin{align*} P(L)\times 1000=0.74\times 1000=740 \end{align*}$

c) First we need to find the the probability that a lost hiker is found (event F), given that the hiker doesn't wear a locator (event NL) and it is given by

$\begin{align*} P(F\mid NL)&=\frac{NL\cap F}{P(NL)}\quad\text{using the formula for conditional probabilities}\\ &=\frac{0.10}{0.26}\\ &=0.385 \end{align*}$

Now if we multiply the above by 1000 we will get the required number

$\begin{align*} P(F\mid NL)\times 1000=0.385\times 1000=385 \end{align*}$

ans: From a total of 1000 lost hikers who don't wear a locator, the number expected to be found is 385.

d) If the hiker is found (event F), the probability that the hiker was wearing a locator is same as

the probability that the hiker was wearing a locator (event L) given that the hiker is found (event F)

$\begin{align*} P(L\mid F)&=\frac{L\cap F}{P(F)}\quad\text{using the formula for conditional probabilities}\\ &=\frac{0.70}{0.80}\\ &=\frac{7}{8} \end{align*}$

e) Let us construct the tree diagram with the event if a lost hiker wears locator as the first set of branches

Hiker is found/7 P(FIL)-0.94 P(F,L)=P(L)*P(FIL)=0.70 2 Wear locator P(L)-0.74 Hiker is not found P(NF L)-0.054 P(NF,L)-P(L)*P(NF L)-0.04 1 Does not Wear locator P(NL)=0.26 Hiker is foundP(F,NL)-P(NL)*P(F NL)-0.10 P(FINL)-0.385 Hiker is not found P(NFINL)-0.615 P(NF,NL)-P(NLy P(NFINL)-0.16

Node 1: We know the marginal probability that a lost hiker would be wearing a locator is P(L)=0.74

The marginal probability that a lost hiker would not be wearing a locator is P(NL)=0.26

Node 2: Given that the lost hiker is wearing a locator, the probability that a hiker is found is

$\begin{align*} P(F\mid L)&=\frac{L\cap F}{P(L)}\quad\text{using the formula for conditional probabilities}\\ &=\frac{0.70}{0.74}\\ &=0.946 \end{align*}$

Given that the lost hiker is wearing a locator, the probability that a hiker is not found is

$\begin{align*} P(NF\mid L)&=\frac{L\cap NF}{P(L)}\quad\text{using the formula for conditional probabilities}\\ &=\frac{0.04}{0.74}\\ &=0.054 \end{align*}$

Node 3: Given that the lost hiker is not wearing a locator, the probability that a hiker is found is

$\begin{align*} P(F\mid NL)&=\frac{NL\cap F}{P(NL)}\quad\text{using the formula for conditional probabilities}\\ &=\frac{0.10}{0.26}\\ &=0.385 \end{align*}$

Given that the lost hiker is not wearing a locator, the probability that a hiker is not found is

$\begin{align*} P(NF\mid NL)&=\frac{NL\cap NF}{P(NL)}\quad\text{using the formula for conditional probabilities}\\ &=\frac{0.16}{0.26}\\ &=0.615 \end{align*}$