Question

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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

(a)
P(Locator & Not found) = 0.04

EXPLANATION: In the Table given, for Column = Locator & Column = Not found, probability = 0.04

(b)

P(Locator) = 0.74

EXPLANATION: In the Table given, the column total for Locator = 0.74

So,

Expected number of hikers with Locator = 1000 X 0.74 = 740

(c)

P(Found/No locator) = P(Found & No locator) / P(No locator)

= 0.10/0.26

= 0.3846

So,

Out of 1000 No Locator, expected number of found = 1000 X 0.3846 = 385

(d)

P(Locator/ Found) = P(Locator & Found)/ P(Found)

= 0.70/0.80 = 7/8

As per directions for answering, first 4 questions are answered.