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Two-ships-have-the-same-berth-in-a-port-It-is-known-that-the-arrival-times-of-the-two-ships-are-independent-and-have-the-same-probability-of-docking-on-a-Sunday-00-00-24-00-If-the-berth-

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Question Number 207752 by efronzo1 last updated on 25/May/24
Two ships have the same berth in a port. It is known that the arrival times of the two ships are independent and have the same probability of docking on a Sunday (00.00−24.00) If the berth time of the first ship is 2 hours and the berth time of the second ship is 4 hours, the probability that one ship will have to wait until the berth can be used is □ ((67)/(144)) □ ((67)/(288)) □ (1/4) □((33)/(144))
$$\:\mathrm{Two}\:\mathrm{ships}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{berth}\: \\ $$$$\:\mathrm{in}\:\mathrm{a}\:\mathrm{port}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\:\mathrm{arrival}\:\mathrm{times}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{ships}\: \\ $$$$\:\mathrm{are}\:\mathrm{independent}\:\mathrm{and}\:\mathrm{have}\:\mathrm{the}\: \\ $$$$\:\mathrm{same}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{docking}\: \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{Sunday}\:\left(\mathrm{00}.\mathrm{00}−\mathrm{24}.\mathrm{00}\right) \\ $$$$\:\mathrm{If}\:\mathrm{the}\:\mathrm{berth}\:\mathrm{time}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{ship} \\ $$$$\:\mathrm{is}\:\mathrm{2}\:\mathrm{hours}\:\mathrm{and}\:\mathrm{the}\:\mathrm{berth}\:\mathrm{time} \\ $$$$\:\mathrm{of}\:\mathrm{the}\:\mathrm{second}\:\mathrm{ship}\:\mathrm{is}\:\mathrm{4}\:\mathrm{hours},\: \\ $$$$\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{one}\:\mathrm{ship} \\ $$$$\:\mathrm{will}\:\mathrm{have}\:\mathrm{to}\:\mathrm{wait}\:\mathrm{until}\:\mathrm{the} \\ $$$$\:\mathrm{berth}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used}\:\mathrm{is}\: \\ $$$$\:\Box\:\frac{\mathrm{67}}{\mathrm{144}}\:\:\:\:\:\Box\:\frac{\mathrm{67}}{\mathrm{288}}\:\:\:\:\Box\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\Box\frac{\mathrm{33}}{\mathrm{144}} \\ $$
Commented by mr W last updated on 25/May/24
p=1−((22^2 +20^2 )/(2×24^2 ))=((67)/(288)) ✓
$${p}=\mathrm{1}−\frac{\mathrm{22}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }{\mathrm{2}×\mathrm{24}^{\mathrm{2}} }=\frac{\mathrm{67}}{\mathrm{288}}\:\checkmark \\ $$
Commented by efronzo1 last updated on 25/May/24
why sir 1−((22^2 +24^2 )/(2.24^2 )) ?
$$\mathrm{why}\:\mathrm{sir}\:\mathrm{1}−\frac{\mathrm{22}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} }{\mathrm{2}.\mathrm{24}^{\mathrm{2}} }\:? \\ $$
Commented by mr W last updated on 25/May/24
my method see below
$${my}\:{method}\:{see}\:{below} \\ $$
Answered by mr W last updated on 25/May/24
Commented by mr W last updated on 25/May/24
say the first ship arrives at time x (0≤x≤24) and the second ship at time y (0≤y≤24). such that the second ship doesn′t need to wait: y≥x+2 ...(i) such that the first ship doesn′t need to wait: x≥y+4 ...(ii) we can intepret this geometrically: each point (x, y) in the square 24×24 shows the random arrival times of both ships. if this point lies in the hatched zones, which mean the conditions (i) and (ii), then no ship needs to wait. otherwise one ship has to wait. the area of hatched zones is ((22^2 +20^2 )/2). the total area is 24^2 . so the probability that one ship has to wait is p=((24^2 −((22^2 +20^2 )/2))/(24^2 ))=1−((22^2 +20^2 )/(2×24^2 ))=((67)/(288))
$${say}\:{the}\:{first}\:{ship}\:{arrives}\:{at}\:{time}\:{x} \\ $$$$\left(\mathrm{0}\leqslant{x}\leqslant\mathrm{24}\right)\:{and}\:{the}\:{second}\:{ship}\:{at} \\ $$$${time}\:{y}\:\left(\mathrm{0}\leqslant{y}\leqslant\mathrm{24}\right). \\ $$$${such}\:{that}\:{the}\:{second}\:{ship}\:{doesn}'{t} \\ $$$${need}\:{to}\:{wait}: \\ $$$${y}\geqslant{x}+\mathrm{2}\:\:\:\:\:…\left({i}\right) \\ $$$${such}\:{that}\:{the}\:{first}\:{ship}\:{doesn}'{t} \\ $$$${need}\:{to}\:{wait}: \\ $$$${x}\geqslant{y}+\mathrm{4}\:\:\:…\left({ii}\right) \\ $$$${we}\:{can}\:{intepret}\:{this}\:{geometrically}: \\ $$$${each}\:{point}\:\left({x},\:{y}\right)\:{in}\:{the}\:{square} \\ $$$$\mathrm{24}×\mathrm{24}\:{shows}\:{the}\:{random}\:{arrival} \\ $$$${times}\:{of}\:{both}\:{ships}.\:{if}\:{this}\:{point}\:{lies} \\ $$$${in}\:{the}\:{hatched}\:{zones},\:{which}\:{mean} \\ $$$${the}\:{conditions}\:\left({i}\right)\:{and}\:\left({ii}\right),\:{then} \\ $$$${no}\:{ship}\:{needs}\:{to}\:{wait}.\:{otherwise}\: \\ $$$${one}\:{ship}\:{has}\:{to}\:{wait}. \\ $$$${the}\:{area}\:{of}\:{hatched}\:{zones}\:{is} \\ $$$$\frac{\mathrm{22}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }{\mathrm{2}}.\:{the}\:{total}\:{area}\:{is}\:\mathrm{24}^{\mathrm{2}} . \\ $$$${so}\:{the}\:{probability}\:{that}\:{one}\:{ship} \\ $$$${has}\:{to}\:{wait}\:{is} \\ $$$${p}=\frac{\mathrm{24}^{\mathrm{2}} −\frac{\mathrm{22}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }{\mathrm{2}}}{\mathrm{24}^{\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{22}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }{\mathrm{2}×\mathrm{24}^{\mathrm{2}} }=\frac{\mathrm{67}}{\mathrm{288}} \\ $$

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