Student#123
Badges: 6
Rep:
?
#1
Report Thread starter 8 years ago
#1
hi im doing an application and it asks me this question...

Three fair dice are thrown. What is the probability the exactly two of the scores
are sixes?

On my way to school I pass through two sets of traffic lights that operate
independently. The probabilities that I have to wait at these two sets of traffic
lights are 0.3 and 0.4 respectively. What is the probability that I am delayed by at least one of the sets of traffic lights?

A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
outcome?



im not good at this type of maths so please help me clearly.

thank you!
0
reply
SubAtomic
Badges: 11
Rep:
?
#2
Report 8 years ago
#2
(Original post by Student#123)
...
What have you tried?

Did you make the questions up or is that how they are written?
0
reply
SubAtomic
Badges: 11
Rep:
?
#3
Report 8 years ago
#3
Do you mean roll three dice and you get a double six?

P(6) P(6') does this mean anything to you?
0
reply
Student#123
Badges: 6
Rep:
?
#4
Report Thread starter 8 years ago
#4
(Original post by SubAtomic)
Do you mean roll three dice and you get a double six?

P(6) P(6') does this mean anything to you?
Three fair dice are thrown. What is the probability the exactly two of the scores
are sixes?


For this question I did

 P(exactly two sixes)=P(6) \times P(6)   \timesP(6')=(\frac{1}{6})^2 \times \frac{5}{6}=\frac{5}{216} which is incorrect.


On my way to school I pass through two sets of traffic lights that operate
independently. The probabilities that I have to wait at these two sets of traffic
lights are 0.3 and 0.4 respectively. What is the probability that I am delayed by at least one of the sets of traffic lights?

P(delay on 1st and 2nd)=0.3 \times 0.4=0.12

p(delayed by at least 1 traffic light)=1-(0.12)^2=0.9856 which is incorrect.


A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
outcome?

I attempted a tree diagram for this question but it got messy and confused me.
0
reply
nawfaall
Badges: 0
Rep:
?
#5
Report 8 years ago
#5
1/6*1/6 if Im not mistaken


Posted from TSR Mobile
1
reply
Student#123
Badges: 6
Rep:
?
#6
Report Thread starter 8 years ago
#6
(Original post by nawfaall)
1/6*1/6 if Im not mistaken


Posted from TSR Mobile
that's wrong too.
0
reply
Mullah.S
Badges: 0
Rep:
?
#7
Report 8 years ago
#7
the answer is


3*5/(6^3)
0
reply
SubAtomic
Badges: 11
Rep:
?
#8
Report 8 years ago
#8
(Original post by Student#123)
Three fair dice are thrown. What is the probability the exactly two of the scores
are sixes?


For this question I did

 P(exactly two sixes)=P(6) \times P(6)-P(6')=(\frac{1}{6})^2 \times \frac{5}{6}=\frac{5}{216} which is incorrect.
Why did you do - P(6') ? The three dice will show something so it is multiply all the way. P(6) and P(6) and P(6'). 5/216 is correct. Unless it is asking for the sum of two dice to be six :s. The question is badly written.

(Original post by Student#123)
On my way to school I pass through two sets of traffic lights that operate
independently. The probabilities that I have to wait at these two sets of traffic
lights are 0.3 and 0.4 respectively. What is the probability that I am delayed by at least one of the sets of traffic lights?

P(delay on 1st and 2nd)=0.3 \times 0.4=0.12

p(delayed by at least 1 traffic light)=1-(0.12)^2=0.9856[/tex] which is incorrect.
Add up all the possibilities, both traffic lights could delay you, or one could or the other could.


(Original post by Student#123)
A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
outcome?

I attempted a tree diagram for this question but it got messy and confused me.
P(HHH)= (0.47)^3

Does that help?
0
reply
Student#123
Badges: 6
Rep:
?
#9
Report Thread starter 8 years ago
#9
(Original post by SubAtomic)
Why did you do - P(6') ? The three dice will show something so it is multiply all the way. P(6) and P(6) and P(6'). 5/216 is correct.



Add up all the possibilities, both traffic lights could delay you, or one could or the other could.




P(HHH)= (0.47)^3

Does that help?
5/216 is not correct.

I have to choose from:

(b) 5/72
(c) 25/72
(d) 25/216




. A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
outcome?
(a) Two tails and one head
(b) Two heads and one tail
(c) Three heads
(d) Three tails
0
reply
Mullah.S
Badges: 0
Rep:
?
#10
Report 8 years ago
#10
(Original post by Student#123)
hi im doing an application and it asks me this question...

Three fair dice are thrown. What is the probability the exactly two of the scores
are sixes?

On my way to school I pass through two sets of traffic lights that operate
independently. The probabilities that I have to wait at these two sets of traffic
lights are 0.3 and 0.4 respectively. What is the probability that I am delayed by at least one of the sets of traffic lights?

A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
outcome?



im not good at this type of maths so please help me clearly.

thank you!

the answer is being 3(5)/6^3

there are 6^3 total possible out come

there are three possible outcome of 6, 6, non six

each non six can be five possibility

so answer is 15/216




EDIT: 15/216 = 5/72
2
reply
Student#123
Badges: 6
Rep:
?
#11
Report Thread starter 8 years ago
#11
(Original post by Mullah.S)
the answer is being 3(5)/6^3

there are 6^3 total possible out come

there are three possible outcome of 6, 6, non six

each non six can be five possibility

so answer is 15/216
thank you
0
reply
Mullah.S
Badges: 0
Rep:
?
#12
Report 8 years ago
#12
for second question:



0.3 and 0.4 probability delay


total options are equal:


delay - delay (0.3*0.4)

delay - non delay (0.3*0.6)

non delay - delay (0.7*0.4)

non delay - non delay (0.7*0.6)


total probability of 1 delay = (0.3*0.6)+(0.7*0.4) / (0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)
0
reply
Student#123
Badges: 6
Rep:
?
#13
Report Thread starter 8 years ago
#13
(Original post by SubAtomic)
Why did you do - P(6') ? The three dice will show something so it is multiply all the way. P(6) and P(6) and P(6'). 5/216 is correct. Unless it is asking for the sum of two dice to be six :s. The question is very badly written.



Add up all the possibilities, both traffic lights could delay you, or one could or the other could.




P(HHH)= (0.47)^3

Does that help?
I get 0.53 for the traffic light question.

A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
outcome?
(a) Two tails and one head
(b) Two heads and one tail
(c) Three heads
(d) Three tails
(e) I don’t know

I thought 3 tails since this has larger probability bit is wrong... could you explain please?
0
reply
Student#123
Badges: 6
Rep:
?
#14
Report Thread starter 8 years ago
#14
(Original post by Mullah.S)
for second question:



0.3 and 0.4 probability delay


total options are equal:


delay - delay (0.3*0.4)

delay - non delay (0.3*0.6)

non delay - delay (0.7*0.4)

non delay - non delay (0.7*0.6)


total probability of 1 delay = (0.3*0.6)+(0.7*0.4) / (0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)

I have answer as 0.58

A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
outcome?
(a) Two tails and one head (b) Two heads and one tail
(c) Three heads (d) Three tails
(e) I don’t know
0
reply
SubAtomic
Badges: 11
Rep:
?
#15
Report 8 years ago
#15
(Original post by Student#123)
5/216 is not correct.

I have to choose from:

(b) 5/72
(c) 25/72
(d) 25/216
Assuming is fraught with peril. 3 fair dice are thrown and the sum of exactly two dice is six then.
0
reply
Mullah.S
Badges: 0
Rep:
?
#16
Report 8 years ago
#16
3rd question:


possible outcomes:

three H (1 time), two H one T (3 times), two T one H (3 times), three T (1 time)

probabilities

0.47^3 or 3(0.47^2)(0.53) or 3(0.53^2)(0.47) or 0.53^3


which one of this 4 number is being biggest number?
0
reply
Mullah.S
Badges: 0
Rep:
?
#17
Report 8 years ago
#17
(Original post by SubAtomic)
Whoever wrote that question needs a slap. Either 3 fair dice are thrown and the sum is 6 or the sum of two dice is six then.





You'll need to go through the possible outcomes using the method I showed. However it should be obvious.

 P(HHT) = P(THH) \neq P(TTH)

you are needing to learn some amateur statistic (like me). do looking at my answer for correction.
0
reply
Mullah.S
Badges: 0
Rep:
?
#18
Report 8 years ago
#18
(Original post by Student#123)
I have answer as 0.58

A biased coin with P(H) = 0.47 is tossed three times. What is the most likely
outcome?
(a) Two tails and one head (b) Two heads and one tail
(c) Three heads (d) Three tails
(e) I don’t know

you need to put this in calculator

(0.3*0.6)+(0.7*0.4) / (0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)
1
reply
Student#123
Badges: 6
Rep:
?
#19
Report Thread starter 8 years ago
#19
(Original post by Mullah.S)
you are needing to learn some amateur statistic (like me). do looking at my answer for correction.
i understand this question now. thank you.

is 0.58 correct for my 2nd question i post?
0
reply
Student#123
Badges: 6
Rep:
?
#20
Report Thread starter 8 years ago
#20
(Original post by Mullah.S)
you need to put this in calculator

(0.3*0.6)+(0.7*0.4) / (0.3*0.4)+(0.3*0.6)+(0.7*0.4)+(0.7*0.6)
ok. what formula did you use?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Poll: What factors affect your mental health most right now? Post-lockdown edition

Anxiousness about restrictions easing (30)
5.23%
Uncertainty around my education (67)
11.67%
Uncertainty around my future career prospects (66)
11.5%
Lack of purpose or motivation (75)
13.07%
Lack of support system (eg. teachers, counsellors, delays in care) (33)
5.75%
Impact lockdown had on physical health (30)
5.23%
Social worries (incl. loneliness/making friends) (61)
10.63%
Financial worries (36)
6.27%
Concern about myself or my loves ones getting/having been ill (23)
4.01%
Exposure to negative news/social media (34)
5.92%
Difficulty accessing real life entertainment (15)
2.61%
Lack of confidence in making big life decisions (57)
9.93%
Worry about missed opportunities during the pandemic (47)
8.19%

Watched Threads

View All