help me with this maths question!

Original post by Student#123

...

What have you tried?

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

Reply 2

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

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

(edited 11 years ago)

Reply 3

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

Unparseable latex formula:

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.

(edited 11 years ago)

Reply 4

nawfaall

1/6*1/6 if Im not mistaken

Posted from TSR Mobile

Reply 5

Original post by nawfaall

1/6*1/6 if Im not mistaken

Posted from TSR Mobile

that's wrong too.

Reply 6

the answer is

3*5/(6^3)

(edited 11 years ago)

Reply 7

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.

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

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?

(edited 11 years ago)

Reply 8

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

(edited 11 years ago)

Reply 9

Original post by Student#123

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

(edited 11 years ago)

Reply 10

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

(edited 11 years ago)

Reply 11

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)

Reply 12

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? :smile:

Reply 13

Original post by Mullah.S

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

Reply 14

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.

(edited 11 years ago)

Reply 15

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?

(edited 11 years ago)

Reply 16