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))


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Question Number 207752 by efronzo1 last updated on 25/May/24

$Two ships have the same berth$ $in a port . It is known that the$ $arrival times of the two ships$ $are independent and have the$ $same probability of docking$ $on a Sunday (00.00 - 24.00)$ $If the berth time of the first ship$ $is 2 hours and the berth time$ $of the second ship is 4 hours,$ $the probability that one ship$ $will have to wait until the$ $berth can be used is$ $◻ \frac{67}{144} ◻ \frac{67}{288} ◻ \frac{1}{4} ◻ \frac{33}{144}$
Commented by mr W last updated on 25/May/24

$p = 1 - \frac{22^{2} + 20^{2}}{2 \times 24^{2}} = \frac{67}{288} ✓$
Commented by efronzo1 last updated on 25/May/24

$why sir 1 - \frac{22^{2} + 24^{2}}{2 . 24^{2}} ?$
Commented by mr W last updated on 25/May/24

$m y m e t h o d s e e b e l o w$
	Answered by mr W last updated on 25/May/24

	Commented by mr W last updated on 25/May/24

	$s a y t h e f i r s t s h i p a r r i v e s a t t i m e x$ $(0 ⩽ x ⩽ 24) a n d t h e s e c o n d s h i p a t$ $t i m e y (0 ⩽ y ⩽ 24) .$ $s u c h t h a t t h e s e c o n d s h i p {d o e s n}^{'} t$ $n e e d t o w a i t :$ $y ⩾ x + 2 . . . (i)$ $s u c h t h a t t h e f i r s t s h i p {d o e s n}^{'} t$ $n e e d t o w a i t :$ $x ⩾ y + 4 . . . (i i)$ $w e c a n i n t e p r e t t h i s g e o m e t r i c a l l y :$ $e a c h p o i n t (x, y) i n t h e s q u a r e$ $24 \times 24 s h o w s t h e r a n d o m a r r i v a l$ $t i m e s o f b o t h s h i p s . i f t h i s p o i n t l i e s$ $i n t h e h a t c h e d z o n e s, w h i c h m e a n$ $t h e c o n d i t i o n s (i) a n d (i i), t h e n$ $n o s h i p n e e d s t o w a i t . o t h e r w i s e$ $o n e s h i p h a s t o w a i t .$ $t h e a r e a o f h a t c h e d z o n e s i s$ $\frac{22^{2} + 20^{2}}{2} . t h e t o t a l a r e a i s 24^{2} .$ $s o t h e p r o b a b i l i t y t h a t o n e s h i p$ $h a s t o w a i t i s$ $p = \frac{24^{2} - \frac{22^{2} + 20^{2}}{2}}{24^{2}} = 1 - \frac{22^{2} + 20^{2}}{2 \times 24^{2}} = \frac{67}{288}$
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