Hey there! Sign in to join this conversationNew here? Join for free
 You are Here: Home >< Maths

# help me with this maths question! Watch

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!
2. (Original post by Student#123)
...
What have you tried?

Did you make the questions up or is that how they are written?
3. Do you mean roll three dice and you get a double six?

P(6) P(6') does this mean anything to you?
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

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?

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.
5. 1/6*1/6 if Im not mistaken

Posted from TSR Mobile
6. (Original post by nawfaall)
1/6*1/6 if Im not mistaken

Posted from TSR Mobile
that's wrong too.
7. the answer is

3*5/(6^3)
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

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(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.

Does that help?
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.

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
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
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
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)
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.

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?
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
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.
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?
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.

you are needing to learn some amateur statistic (like me). do looking at my answer for correction.
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)
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?
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?

Reply
Submit reply
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 17, 2013
Today on TSR

### Last-minute PS help

100s of personal statements examples here

### Loneliness at uni

Discussions on TSR

• Latest
• ## 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.

• Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## 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.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.