🎓 Academic Mga boss, baka may magaling jan sa inyo sa math, patulong naman po nito:

[electromagnus06]

Suppose that 70% of drivers are classed as being “careful” whereas 30% of the drivers are classed as being “reckless”. Suppose further that a careful driver has a 0.05 probability of being in an accident in a given year, while for a reckless driver the probability is 0.2. What is the probability that a randomly selected driver will have an accident within a year?

 

[K i R O]

0.75

 

[inusena]

try laang
(0.7 * 0.05) + (0.3 * 0.2)
= 0.035 + 0.06
= 0.095 or 9.5%

 

[electromagnus06]

0.75

Pano po ang solution?

 

[Syenderr]

0.095 percent

 

[Syenderr]

Probabilty of a careful driver to have accident = 70 percent x (0.05)
Probabilty of a reckless driver to have accident = 30 percent x (.02)

Add both probabilities = .095 %

 

[electromagnus06]

ang nasa solution manual po kasi ay 75% pero di ko po alam ang solution kung paano.

 

[inusena]

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[Chuckedoll]

C=70% ,(70%x.05%) , 70%x.2
=.035 , .14
R=30% ,(30%x.05%) , 30%x.2
=.015 , .06
add all =.25%
100-25= .75

 

[Chuckedoll]

Its all about probability
there are 75% random driver can be on accident each year
C&R are all drivers

 

[electromagnus06]

Marami pong salamat mga boss.

 

[Syenderr]

charot lang pala haha

 

[inusena]

labo nung 75% pano nangyari yun

 

[electromagnus06]

kasi random siguro?

 

[inusena]

ano final answer mo?

 

[inusena]

sa link na binigay ko 9.5% ang sagot, baka mali ang intindi nila kasi yung answer sa baba " Answers to problems on other side "

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[Chuckedoll]

parang ganito lang kasi yan
may 3 choices ka
each have 33.33%
ngaun tinagal mo ang isa which dalawa nalang choices
ilang percent nung dalawa?
mostly answered it 50/50 but not
its 66.66% vs 33.33%

 

[Zhalamand3rz]

sakit sa ulo nyan.. hahahaha

 

[K i R O]

na DM ko na sayo lods

 

[ThreadRipper]

Kinalawang na utak ko, lulupet ng may mga sagot. Salute!