A maths probability question- please help!!?

Question

Consider this scenario-
Vehicles approaching a crossroads from a particular direction encounter traffic lights that r either red or green.Each vehicle then either goes left,right or straight on.The timing cycle 4 the lights is 1min on red alternating with 45secs on green.A person watching vehicles approaches the junction 4 a few mins&counts 6 that turn left,3 that turn right and 11 that go straight on.

a) use this timing information for the lights to calculate the theoratical probabilities of finding them at red or green
b)use the vehicle count data to calculate estimated probabilities 4a vehicle going left, right or straight on at the junction
c)4 the next vehicle approaching the junction draw a tree diagram showing all the possible combined outcomes with their respective probabilities, use ur diagram2answer d to f

What is the probability that the next vehicle will-
d, find the light on green&go straight on
e,not go straight on
f,find the lights on red & not turn right ???

Answer 1

Answers
a) p(red) = 60sec / (60 + 45) sec = 0.57
p(green) = 45 / 105 s = 0.43

b) p (left) = 6 / 20 (total no of vehicles seen) = 0.3
p (straight on) = 11/20 etc etc

c) Draw the tree diagram yourself

d) p (green light; straight on) = 0.43 x 0.55 = 0.237

e) p (not go straight on) = 1 - 0.55 = 0.45

f) p (red lights, not turn right) = 0.57 x (1- 0.15) = 0.48

Oh and next time do your own maths prep.

Answer 2

Well the light cycle is 105 seconds, so the probability of it being red is 60/105 = 4/7. Similarly the chances of it being green are 45/105 = 3/7.

Out of 20 cars, 6 went left, 3 right and 11 straight, so based on that the chances are 6/20 (= 3/10), 3/20 and 11/20 respectively.

Can't draw you a diagram here though bach, sorry! I can explain what you need to do if you want - let me know.

(d) The chance of green and straight is 3/7 x 11/20 = 33/140.

(e) Not straight (ie left or right) = 9/20

(f) Red and not right (ie left or straight) is 4/7 x 17/20 = 68/140 = 17/35

Answer 3

This could be complicated for you - there are distinctive diverse solutions right here and all people thinks theirs makes appropriate experience. Intuitively, the respond could be around 15% as Euphojim stated. this is purely logical, good? Statistically or mathematically, it does not somewhat paintings like that. Cccccrazy's answer replaced into the main fabulous answer. The question asks for the threat that precisely a million affected person has a reaction. this suggests that 14 sufferers don't get a reaction. So the threat that Joe (random affected person) gets a reaction is a million% and the threat that Jack, James, etc don't get a reaction is ninety 9% for each of them. once you paintings out percentages for distinctive somewhat some issues on the same time (like who has a reaction and who does not), you multiply the guy percentages. this suggests that the possibilities of Joe getting sick and noone else is a million%*ninety 9%*ninety 9%*ninety 9%*ninety 9%*ninety 9%*ninety 9%*ninety 9%*ninety 9%*ninety 9%*9... or (0.01)^a million * (0.ninety 9)^14 = 0.008687 yet then the prospect that Jack gets sick and all people else remains the same is 0.008687, and the same for each guy or woman else. So, for all 15 of them, the entire threat is the sum of each and every individual's threat of having sick, or: 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 + 0.008687 = 15 * (0.008687) = 0.1303 =13.03% wish that helps make it sparkling. once you get into information slightly added, you will see that this would be a appropriate occasion of a Binomial Distribution. then you definitely can purely pop the numbers right into a formulation so this is much less stressful.

Answer 4

The light is red for 60 seconds, green for 45 seconds. Whole cylce is 105 seconds.
P(red) = 60/105

for a) you need only know the rate at which cars are approaching. If their arrival is truly random, then the probability of a car arriving at green is 1-P(red).

The fraction of cars turning appears to be independent of the light color. therefore, P(turn_left)= 6/20

Therefore, the probability that a car arrives on red AND turns left should be
P(red)*P(turn_left)

The probability of red light AND not turn right =
P(red)*(1-P(turn_right))

Answer 5

a) red 4/7 Green 3/7
b)left 3/10 right 3/20 straight 11/20
c)no
d)33/140
e)9/20
f)17/35

Answer 6

If The Dice Is On A Three, It Is Luckily To Turn To A Eight If You Move It Two Iches Away From You.