calculate the probability?

Question

on the way to college each morning susan has to pass though three sets of traffic lights in the city centre. the probalities that she passes through them without having to stop are 0.3, 0.5 and 0.6respectively. you may assume that each set of lights operates independently of the others.

(a)Find the probability that susan does not have to stop at any of the three lights.

(b)Find the probability that she has to stop at just one set of lights

(c)given that she has to stop at just one set of lights,find the probability that she has to stop at first set of lights
it is decided to change the probability of passing through the first light without having to stop from 0.3 to p. susan finds that his change reduces the probability found in(c) to 0.5. calculate p.

Answer 1

a) 46% chance
b)30%chance
c)42%

Answer 2

a. 9/100
b. 9/25
c.7/12
p=4/10
i'm pretty sure it is 4/10 because if you look at c the answer is given as 21/36 or 7/12 if reduced. to get .5 you need to make it 18/36 so then u have equation (1-p)(5/10)(6/10)=18/100.
(1-p)(5/10)=(3/10). then you have 1-p=3/5 or 1-p=6/10 so p=4/10

Answer 3

a) 0.3*0.5*0.6 (you MULTIPLY independent events along a path) = 0.09 - not too likely.

b) one set, could be any set, so we have 3 different paths to SUM up.
path 1: stops at first set: 0.7*0.5*0.6 = 0.21
path 2: stops at second set: 0.3*0.5*0.6 = 0.09
path 3: stops at third set: 0.3*0.5*0.4 = 0.06
summing these up: 0.36

c) the probability of stopping at 1 set of lights is .36, from b. the probability that it is the first set is .21, also from b. so we have 0.21/0.36 or 7/12.

d) ((1-p)(.30)) / ((1-p)(.30)+(p)(.30)+(p)(.20)) = 0.5
(.30-.30p) / (.30-.30p+.50p) = 0.5
(.30-.30p) / (.30+.20p) = 0.5
(.60-.60p) / (.30+.20p) = 1
.60 - .60p = .30 + .20p
.30 = .80p
p = 3/8 = .375

Answer 4

I think this is what you are looking for :
(a)
(0∙3)(0∙5)(0∙6) = 0∙09 = 9%

(b)
(1/3)(0∙09) = 0∙03 = 3%

(c)
(0∙3)/ [(0∙3)+(0∙5)+(0∙6)] = 0∙3/ 1∙4 = 0∙214 285 714... ≈ 21∙43%

(p)/ [(0∙3)+(0∙5)+(0∙6)] = 0∙5
(p) = (0∙5)[(0∙3)+(0∙5)+(0∙6)]
(p) = (0∙15)+(0∙25)+(0∙3)
(p) = 0∙7

Answer 5

Doesn't the above set of answers perfectly illustrate the point I have made several times about the number of wrong answers that appear on this website? Partly for this reason, students should not copy answers, but use help given on how to answer the question to do it themselves.

Answer 6

For a), that would just be 0.3*0.5*0.6 = 0.09.

For b), that would be 0.7*0.5*0.6 + 0.3*0.5*0.6 + 0.3*0.5*0.4 = 0.21 + 0.09 + 0.06 = 0.36.

For c), that would be (0.7*0.5*0.6)/(0.36) = 0.58

If 0.3 was changed to p, then c) would be solved through the following way:

(p*0.5*0.6)/((1-p)*0.5*0.6 + p*0.5*0.6 + p*0.5*0.4) = 0.5
0.3p/(0.3 - 0.3p + 0.3p + 0.2p) = 0.5
0.3p/(0.3 + 0.2p) = 0.5
0.3p = 0.5(0.3 + 0.2p)
0.3p = 0.15 + 0.1p
0.2p = 0.15
p = 0.15/0.2
p = 0.75

Answer 7

a) 0.3 * 0.5 * 0.6
b) 0.7 * 0.5 * 0.6 + 0.3 * 0.5 * 0.6 + 0.3 * 0.5 * 0.4
c) ( 0.7 * 0.5 * 0.6 ) / ( 0.7 * 0.5 * 0.6 + 0.3 * 0.5 * 0.6 + 0.3 * 0.5 * 0.4 )
d) ( p * 0.5 * 0. 6 ) / ( ( 1 - p ) * 0.5 * 0.6 + p * 0.5 * 0.6 + p * 0.5 * 0.4 ) = 0.5, Solve the equation for p.