Mathematics: Post your doubts here!

[Suchal Riaz]

the answers given do not follow the required proper technique to solve S1 questions, however they seem to be correct except the last part

last one you mean cost? i just wanted to help. i am at the As level only. i have 'officially' only done P1 in school and i am doing P3, M2 and S1 questions just for helping. if you don't get it, i am wrong or it offends you in any way, I am very sorry.

 

[Suchal Riaz]

I just solved the first part in a detailed way but i saw that you wanted the second part. I did it in a rough way. If u don't get it, tell me i will elaborate it and write on a neat paper..

• http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_32.pdf no 8ii

 

[RadzMau]

i have not done S1. i will use my o level knowledge to do it.
(i) if anyone wins the first one, the probability of the same person winning the game is clearly 1/2
(ii) there's only one way this can happen. the one who wins the first one loses the second and third. so it is 1/2 * 1/2 = 1/4
(iii) i guess there is again only one possibility: first he wins, second loses, third wins and fourth wins. so 1/2 * 1/2 * 1/2 = 1/8
(iv) well win lose win lose win is the only possibility as any other way the game will end just there. it will be by sequence (1/2)^4

there is no way the game can go on after 5 games.
the cost will be i guess 18/100.

Hey, I appreciate your effort! But S1 knowledge is required for this question. anyway, thanks

 

[Suchal Riaz]

what an attitude.

Do you have any idea what answering a question takes. I need to solve it, then confirm it. Then write it on a good piece of paper. Take a photo. Upload it here. Then people tell me that they can't even lift their fingers to just send me the link to make the process faster. Like i am a servant of yours. When i am doing so much for others, they can't even post link?
Why would someone do so much for others? I don't know. I just like to do this. But this is how they repay me? You people think we get paid to answer ur questions? I don't even ask questions from others. From last 3 years i have hardly asked any question on XPC. And you know my attitude? Nah. The rules of this forum doesn't allow me to use the kind of language which i use with d*ckheads like you.

 

[Suchal Riaz]

Hey, I appreciate your effort! But S1 knowledge is required for this question. anyway, thanks

I wish i could be of more help.

 

[ahmed abdulla]

View attachment 36719

Well explained ,
one question : all your steps are clear , but still many cases they give same diagram but ADD , give positive to BOTH figures !
thanks

 

[wohooo]

I just solved the first part in a detailed way but i saw that you wanted the second part. I did it in a rough way. If u don't get it, tell me i will elaborate it and write on a neat paper..

View attachment 36742

Thank you

 

[Ahmed Aqdam]

Hello Everyone!
Kindly please solve this Statistics 1 question on probability.
Thanks

If man A wins the first two games, there are 2^3=8 ways the last 3 games could have been played.
If he wins in three games, there are 2^2=4 ways the last 2 games could have been played.
If he wins in four, the last game could only have 2 possibilities.
If he wins in 5 there is only one way it could happen.
In total there are 8+4+2+1=15 ways the match could be played.
The probabilities will be 8/15, 4/15, 2/15 and 1/15.

The expected value the winner receives is (8/15*14)+(4/15*14)+(2/15*18)+(1/15*18)=14.8
The amount every spectator gives is 14.8/100=0.148

 

[Talha Irfan]

last one you mean cost? i just wanted to help. i am at the As level only. i have 'officially' only done P1 in school and i am doing P3, M2 and S1 questions just for helping. if you don't get it, i am wrong or it offends you in any way, I am very sorry.

neither did it offend me nor it looks completely incorrect i just said it lacks proper structure of solving AS S1 questions, and jbtw there is obviously no direct return of this help but have you ever heard of karma!? There are million others who help without any will of return. Just look at that famous khan academy guy

 

[Talha Irfan]

If man A wins the first two games, there are 2^3=8 ways the last 3 games could have been played.
If he wins in three games, there are 2^2=4 ways the last 2 games could have been played.
If he wins in four, the last game could only have 2 possibilities.
If he wins in 5 there is only one way it could happen.
In total there are 8+4+2+1=15 ways the match could be played.
The probabilities will be 8/15, 4/15, 2/15 and 1/15.

The expected value the winner receives is (8/15*14)+(4/15*14)+(2/15*18)+(1/15*18)=14.8
The amount every spectator gives is 14.8/100=0.148

if man A wins the first two games, then should not the match end, i mean then there will be no third game right?

 

[Ahmed Aqdam]

if man A wins the first two games, then should not the match end, i mean then there will be no third game right?

There will be no third game but we just suppose that the match was of full length of 5 games. Then man A would win any match in he would win the first two games regardless of what happens next. Its just a way of calculating probability and not what actually happens.

 

[Talha Irfan]

There will be no third game but we just suppose that the match was of full length of 5 games. Then man A would win any match in he would win the first two games regardless of what happens next. Its just a way of calculating probability and not what actually happens.

we have to calculate the probability of what actually happens as probability is all about the actual scene that's going on which we have to visualise and condense in a mathematical form

 

[Talha Irfan]

Hey


Can you please solve this Statistics 1 question on probability? Your help will be much appreciated. Thanks

Possibility of winning case A or B wins
= A wins + B wins

i) Two games
= AA + BB
= P(A) * P(A) + P(B) * P(B) = 1/4 + 1/4 = 1/2
we don't have to include the case in which the match draws.

ii) Three games
= ABB + BAA (the match shouldn't end before third game, i.e it should reach the third game)
= 1/4

similarly
iii) ABAA + BABB = 1/8

and iv) ABABA + BABAB = 1/16

mean cost
p(2 or 3) = 0.5 + 0.25 = 0.75
p(4 or 5) = 1/8 + 1/16 = 3/16

E(X) = (0.75*14 + 3/16*18) / 100 spectators

 

[Suchal Riaz]

I never asked for anything in return. And i can not do the community the help which Sal has done.

neither did it offend me nor it looks completely incorrect i just said it lacks proper structure of solving AS S1 questions, and jbtw there is obviously no direct return of this help but have you ever heard of karma!? There are million others who help without any will of return. Just look at that famous khan academy guy

 

[Lyfroker]

are the tangent graphs required for the AS level? n do i need to know the symmetry properties of the graphs???

 

[Ahmed Aqdam]

Possibility of winning case A or B wins
= A wins + B wins

i) Two games
= AA + BB
= P(A) * P(A) + P(B) * P(B) = 1/4 + 1/4 = 1/2
we don't have to include the case in which the match draws.

ii) Three games
= ABB + BAA (the match shouldn't end before third game, i.e it should reach the third game)
= 1/4

similarly
iii) ABAA + BABB = 1/8

and iv) ABABA + BABAB = 1/16

mean cost
p(2 or 3) = 0.5 + 0.25 = 0.75
p(4 or 5) = 1/8 + 1/16 = 3/16

E(X) = (0.75*14 + 3/16*18) / 100 spectators

Probabilities should add up to 1 but your probabilities add up to 15/16.

The possible results for man A are 15 if he wins (W denotes win and L denotes lose):
W-W-W-W-W
W-W-W-W-L
W-W-W-L-W
W-W-L-W-W
W-W-L-L-W
W-W-L-W-L
W-W-W-L-L
W-W-L-L-L
L-W-W-W-W
L-W-W-W-L
L-W-W-L-W
L-W-W-L-L
W-L-W-W-W
W-L-W-W-L
W-L-W-L-W

In these, 8 will result in win in 2 games, 4 in win 3 games, 2 in win in 4 games and 1 in win in 5 games.

 

[Talha Irfan]

Probabilities should add up to 1 but your probabilities add up to 15/16.

The possible results for man A are 15 if he wins (W denotes win and L denotes lose):
W-W-W-W-W
W-W-W-W-L
W-W-W-L-W
W-W-L-W-W
W-W-L-L-W
W-W-L-W-L
W-W-W-L-L
W-W-L-L-L
L-W-W-W-W
L-W-W-W-L
L-W-W-L-W
L-W-W-L-L
W-L-W-W-W
W-L-W-W-L
W-L-W-L-W

In these, 8 will result in win in 2 games, 4 in win 3 games, 2 in win in 4 games and 1 in win in 5 games.

this is one of the typical questions of probability but the remaining probability which makes my total probability equals to one is of the case in which the match draws when 2/3/4/5 matches are played and also we have to include the probability of both players winning

 

[ShreeyaBeatz]

Guys I need help on this!

 

[ShreeyaBeatz]

Probabilities should add up to 1 but your probabilities add up to 15/16.

The possible results for man A are 15 if he wins (W denotes win and L denotes lose):
W-W-W-W-W
W-W-W-W-L
W-W-W-L-W
W-W-L-W-W
W-W-L-L-W
W-W-L-W-L
W-W-W-L-L
W-W-L-L-L
L-W-W-W-W
L-W-W-W-L
L-W-W-L-W
L-W-W-L-L
W-L-W-W-W
W-L-W-W-L
W-L-W-L-W

In these, 8 will result in win in 2 games, 4 in win 3 games, 2 in win in 4 games and 1 in win in 5 games.

Ahmed help me on this ! PLEASE

 

[Rameesha_a*]

Geometry.. does anyone have easy explaination for geometry if so please help .. doing IGCSE'14 Maths Extended .