Mathematics: Post your doubts here!

[Lee Qian Yi]

I have some questions about statistics.
1) A cubical dice is biased so that the probability of any particular score between 1 and 6 (inclusive) being obtained is proportional to that score. Find the probability of scoring a 1.
ANS :1/21
2) For a biased cubical dice the probability of any particular score between 1 and 6 (inclusive) being obtained is inversely proportional to that score. Find the probability of scoring a 1.
ANS: 20/49
3) A fair coin is tossed 8 times. Calculate the probability that the first 4 tosses and the last 4 tosses result in the same number of heads.
ANS :0.273
4) Two teams each consist of 3 players. Each player in a team tosses a fair coin once and the team score's is the total number of heads thrown. Find the probability that the teams have the same score.
ANS: 0.313

Can someone explain to me how to find the answer ?

 

[Holmes]

9709/12 June 2012 Pure Mathematics !
Help me please!
thanks.

 

[Lee Qian Yi]

View attachment 61874

9709/12 June 2012 Pure Mathematics !
Help me please!
thanks.

Screen Shot 2017-03-29 at 7.09.13 PM

 

[Lee Qian Yi]

View attachment 61874

9709/12 June 2012 Pure Mathematics !
Help me please!
thanks.

 

[DragonCub]

View attachment 61874

9709/12 June 2012 Pure Mathematics !
Help me please!
thanks.

 

[DragonCub]

I have some questions about statistics.
1) A cubical dice is biased so that the probability of any particular score between 1 and 6 (inclusive) being obtained is proportional to that score. Find the probability of scoring a 1.
ANS :1/21
2) For a biased cubical dice the probability of any particular score between 1 and 6 (inclusive) being obtained is inversely proportional to that score. Find the probability of scoring a 1.
ANS: 20/49
3) A fair coin is tossed 8 times. Calculate the probability that the first 4 tosses and the last 4 tosses result in the same number of heads.
ANS :0.273
4) Two teams each consist of 3 players. Each player in a team tosses a fair coin once and the team score's is the total number of heads thrown. Find the probability that the teams have the same score.
ANS: 0.313

Can someone explain to me how to find the answer ?

 

[Lee Qian Yi]

View attachment 61878

oh i get it ! really thx very much.

 

[Lee Qian Yi]

Can help me solve this question?
1)A game is played by throwing a fair dice until either a 6 is obtained or 4 throws have been made. The number of 6s obtained in the game is denoted by Y. Find E(Y). ANS: 0.5177
Pls help to explain this question and why the answer is this, thx.
2) The computer allows a maximum of 4 attempts altogether. For each attempt the probability of success is 0.4, independently of all other attempts. The total number of attempts he makes, successful or not, is denoted by X( possible values of X are 1,2,3,4) . Tabulate the probability distribution of X.
ANS : P(1)=0.4 P(2)=0.24 P(3)=0.144 P(4)=0.216
and ya one more question
3)2 spaniels , 2 retrievers and 3 poodles go through to the final. They are placed in line. How many different arrangement of these 7 dogs are there if no poodle is next to another poodle?
ANS: 1440

 

[DragonCub]

Can help me solve this question?
1)A game is played by throwing a fair dice until either a 6 is obtained or 4 throws have been made. The number of 6s obtained in the game is denoted by Y. Find E(Y). ANS: 0.5177
Pls help to explain this question and why the answer is this, thx.
2) The computer allows a maximum of 4 attempts altogether. For each attempt the probability of success is 0.4, independently of all other attempts. The total number of attempts he makes, successful or not, is denoted by X( possible values of X are 1,2,3,4) . Tabulate the probability distribution of X.
ANS : P(1)=0.4 P(2)=0.24 P(3)=0.144 P(4)=0.216
and ya one more question
3)2 spaniels , 2 retrievers and 3 poodles go through to the final. They are placed in line. How many different arrangement of these 7 dogs are there if no poodle is next to another poodle?
ANS: 1440

(1)
Since each trial either stops when a 6 is obtained, or 4 throws, the number of 6 obtained in each trial can only be 0 or 1.
So the expectancy is the same as the probability of getting a 6 in a trial.
This probability = 1 - (probability of getting no 6 in four throws)
Probability of getting no 6 = (5/6)^4
Therefore: P = 1 - (5/6)^4 = 0.5177

(2)
Similar to (1).
X = 1 means the computer succeeds at the 1st attempt, and X = 2 means success at the 2nd, so on.
So P(1) = 0.4 (i.e. success probability of a single attempt)
P(2) = P(fail 1st, success 2nd) = 0.6 * 0.4 = 0.24 (fail probability * success probability)
P(3) = 0.6 * 0.6 * 0.4 = 0.144
P(4) = 1 - P(1) - P(2) - P(3) = 0.216

(3)
[# of arrangements] = [total # of arrangements] - [# of arrangements where a poodle is next to another]
total # = 7P7 = 7! = 5040

For # of arrangements where a poodle is next to another, we can do the following:
1. Choose 2 poodles from the 3, which means 3C2.
2. Group them into a single "unit", so that instead of 7 dogs, there are now 6 "units" for permutation.
3. Arrange these "units," which means 6P6, or 6!
4. Account for the possible arrangement inside the double-poodle "unit", which has 2 ways
So, the # should be (3C2) * (6!) * 2 = 4320

However, this calculation has actually over-counted the cases where the 3 poodles are all together.
So we need to subtract those cases from the 4320.
How many do we need to subtract?
We can group the 3 poodles together as a "unit", so that there are 5 "units", which means 5P5, or 5!

We also need to include the permutation inside the "unit", which is 3P3, or 3!.
So the number we need to subtract from 4320 is 5! * 3! = 720.

[total # of arrangements] = 7! = 5040
[# of arrangements where a poodle is next to another] = (3C2) * (6!) * 2 - 5! * 3! = 4320 - 720 = 3600
[# of arrangements] = 7! - [(3C2)*6!*2 - 5!*3!] = 1440

 

[Lee Qian Yi]

(1)
Since each trial either stops when a 6 is obtained, or 4 throws, the number of 6 obtained in each trial can only be 0 or 1.
So the expectancy is the same as the probability of getting a 6 in a trial.
This probability = 1 - (probability of getting no 6 in four throws)
Probability of getting no 6 = (5/6)^4
Therefore: P = 1 - (5/6)^4 = 0.5177

(2)
Similar to (1).
X = 1 means the computer succeeds at the 1st attempt, and X = 2 means success at the 2nd, so on.
So P(1) = 0.4 (i.e. success probability of a single attempt)
P(2) = P(fail 1st, success 2nd) = 0.6 * 0.4 = 0.24 (fail probability * success probability)
P(3) = 0.6 * 0.6 * 0.4 = 0.144
P(4) = 1 - P(1) - P(2) - P(3) = 0.216

(3)
[# of arrangements] = [total # of arrangements] - [# of arrangements where a poodle is next to another]
total # = 7P7 = 7! = 5040

For # of arrangements where a poodle is next to another, we can do the following:
1. Choose 2 poodles from the 3, which means 3C2.
2. Group them into a single "unit", so that instead of 7 dogs, there are now 6 "units" for permutation.
3. Arrange these "units," which means 6P6, or 6!
4. Account for the possible arrangement inside the double-poodle "unit", which has 2 ways
So, the # should be (3C2) * (6!) * 2 = 4320

However, this calculation has actually over-counted the cases where the 3 poodles are all together.
So we need to subtract those cases from the 4320.
How many do we need to subtract?
We can group the 3 poodles together as a "unit", so that there are 5 "units", which means 5P5, or 5!

We also need to include the permutation inside the "unit", which is 3P3, or 3!.
So the number we need to subtract from 4320 is 5! * 3! = 720.

[total # of arrangements] = 7! = 5040
[# of arrangements where a poodle is next to another] = (3C2) * (6!) * 2 - 5! * 3! = 4320 - 720 = 3600
[# of arrangements] = 7! - [(3C2)*6!*2 - 5!*3!] = 1440

oh ok thx i get it already but there is no a fix solution for question 3 right? cuz i have see other solution which is



Write down four stars like this:

∗∗∗∗

These represent the positions to be ultimately occupied by the non-poodles. There are 5 gaps , 3 are obvious ones and the 2 are endgames. We must choose 3 of these to be occupied by the poodles. This can be done in( 5p3) ways. The individual poodles can be put in the chosen gaps in 3! orders, and the rest of the dogs can occupy the starred positions in 4! ways, for a total of 5p3*3!*4!

Can u explain why the solution is this?

 

[DragonCub]

oh ok thx i get it already but there is no a fix solution for question 3 right? cuz i have see other solution which is



Write down four stars like this:

∗∗∗∗

These represent the positions to be ultimately occupied by the non-poodles. There are 5 gaps , 3 are obvious ones and the 2 are endgames. We must choose 3 of these to be occupied by the poodles. This can be done in( 5p3) ways. The individual poodles can be put in the chosen gaps in 3! orders, and the rest of the dogs can occupy the starred positions in 4! ways, for a total of 5p3*3!*4!

Can u explain why the solution is this?

↓--↓-↓--↓--↓
_*_*_*_*_
The arrows indicate the 5 gaps to put the poodles in.
You can see that this way, we can ensure that there is at least 1 other breed of dog between two poodles. This is equivalent to "no poodle is next to another poodle."

There is, however, a mistake in the final expression. (5P3) * (3!) * (4!) is wrong (it will get you 8640, not 1440).
Instead, the answer should be (5C3) * (3!) * (4!), or simply (5P3) * (4!).

1. In this solution you got, first step is to choose 3 slots from 5. This means 5C3.
2. Then, we need to put the 3 poodles in those 3 gaps we just chose. This means 3!.

Actually, steps 1 and 2 can be combined together to yield 5P3.

Now that we have finished putting in all the poodles. Then it is:
3. Put the rest 4 dogs in the 4 "*" positions. This means 4!.

Final answer: (5C3) * (3!) * (4!) or (5P3) * (4!) = 1440

There is only one correct final answer to a question, but that doesn't mean there is only one correct way to solve it. And obviously, this way you mentioned is actually more efficient than the one I proposed.

 

[Lee Qian Yi]

↓--↓-↓--↓--↓
_*_*_*_*_
The arrows indicate the 5 gaps to put the poodles in.
You can see that this way, we can ensure that there is at least 1 other breed of dog between two poodles. This is equivalent to "no poodle is next to another poodle."

There is, however, a mistake in the final expression. (5P3) * (3!) * (4!) is wrong (it will get you 8640, not 1440).
Instead, the answer should be (5C3) * (3!) * (4!), or simply (5P3) * (4!).

1. In this solution you got, first step is to choose 3 slots from 5. This means 5C3.
2. Then, we need to put the 3 poodles in those 3 gaps we just chose. This means 3!.

Actually, steps 1 and 2 can be combined together to yield 5P3.

Now that we have finished putting in all the poodles. Then it is:
3. Put the rest 4 dogs in the 4 "*" positions. This means 4!.

Final answer: (5C3) * (3!) * (4!) or (5P3) * (4!) = 1440

There is only one correct final answer to a question, but that doesn't mean there is only one correct way to solve it. And obviously, this way you mentioned is actually more efficient than the one I proposed.

Oh i see thx

 

[krishnapatelzz]

part one in detail :/ .

 

[krishnapatelzz]

q-2 whole

 

[abbas haider]

View attachment 61882part one in detail :/ .

 

[areeba240]

(i)

 

[Holmes]

View attachment 61876

Thanks a lot.

 

[Holmes]

View attachment 61877

Thank you!

 

[DragonCub]

View attachment 61898
(i)

ZERO standard deviation means EVERY value is equal to the mean. So since the mean cost of all of Fei's rides is $2.5, the cost of each ride must also be $2.5.

 

[Saly Elkady]

Do anyone have like a formula sheet or a summary for all the formulas and equations we took in the entire course ? Something to refer to when revising ? ( Pure Maths 1 and 3 , Stats 1 and Mechanics 1 )