# help me with this maths question!

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#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?

thank you! 0
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
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
#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.
0
8 years ago
#5
1/6*1/6 if Im not mistaken

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

Posted from TSR Mobile
that's wrong too.
0
8 years ago
#7

3*5/(6^3)
0
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 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?
0
#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
(d) Three tails
0
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?

thank you! 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

EDIT: 15/216 = 5/72
2
#11
(Original post by Mullah.S)

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

thank you
0
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
#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
(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
#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)

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
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
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
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. you are needing to learn some amateur statistic (like me). do looking at my answer for correction.
0
8 years ago
#18
(Original post by Student#123)

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
#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
#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
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