# help me with this maths question!

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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!

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!

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#2

(Original post by

...

**Student#123**)...

Did you make the questions up or is that how they are written?

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#3

Do you mean roll three dice and you get a double six?

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

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

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(Original post by

Do you mean roll three dice and you get a double six?

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

**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?

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?

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?

outcome?

I attempted a tree diagram for this question but it got messy and confused me.

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#8

(Original post by

For this question I did

which is incorrect.

**Student#123**)**Three fair dice are thrown. What is the probability the exactly two of the scores**

are sixes?are sixes?

For this question I did

which is incorrect.

(Original post by

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

**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?

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.

(Original post by

I attempted a tree diagram for this question but it got messy and confused me.

**Student#123**)**A biased coin with P(H) = 0.47 is tossed three times. What is the most likely**

outcome?

outcome?

I attempted a tree diagram for this question but it got messy and confused me.

Does that help?

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(Original post by

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?

**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?

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

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#10

(Original post by

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!

**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

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(Original post by

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

**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

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#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)

0.3 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)

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(Original post by

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?

**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?

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?

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(Original post by

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)

**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

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#15

(Original post by

5/216 is not correct.

I have to choose from:

(b) 5/72

(c) 25/72

(d) 25/216

**Student#123**)5/216 is not correct.

I have to choose from:

(b) 5/72

(c) 25/72

(d) 25/216

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#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

which one of this 4 number is being biggest number?

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?

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#17

(Original post by

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.

**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.

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#18

(Original post by

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

**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)

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(Original post by

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

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

is 0.58 correct for my 2nd question i post?

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(Original post by

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)

**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)

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