₦airaland Forum

donhils:
Really? Even when Redmi Note 2 couldn't come with ordinary Gorilla screen. What are the good hardware they use pls?

The fact that Redmi note 2, which is just one of their numerous products didn't come with gorilla glass does not make it an inferior product. Moreover, there are other kinds of screen protector apart from gorilla glass in which I'm very sure one was used for the Redmi note 2 as it passed the scratch, drop and water test. So what else are we talking about? BTW, Some of their other products come with gorilla glass too and they're not just called the Apple of China for the sake of it.
We could go on and on about this but I think it's not really worth it. I've made my point, so if you don't agree with it, it's ok.

donhils:
Not the just the name or brand but it's about the entity that Apple represents. It's still Apple phone with their factory in China. It's different from a China brand that's owned and made in China, we all know what China products stand for. A pupil was asked to say the opposite of the word Original and he answered China.
I'm still buying the Xiaomi Redmi Note 2, nothing do me.

My brother, setting sentiments apart, there are lots of these so-called "Chinese" phones that are far better than an iPhone/Samsung products. Xiaomi for instance use good processor and hardware with good designs just as or even better than an iPhone. It's just that people have refused to accept that something good can come out from China and it'll be a difficult task to change their orientation/mentality about it

donhils:
IPhone made in China is different from a China phone.

What makes it different? The name or brand?

Tecno H6
Infinix hot note 2
Tecno camon c7 /C8

And I have a jiayu F2 also for sale at that price range. You can Google out the specs

dhebo:
. ...for guys looking for tutorial in school should try attending this particular one.if Samson from GSF takes u just knw d impact will b great

I'll really like to meet the guy with the great impact, Samson

jackpot:
like solving people's assignments for a fee?

might not necessarily be assignments

Oxytocin:
That's good to hear. Can we hook up??.. and what's your seat number, mine is 48

mine is room 204 seat number 63.

Oxytocin:
Nobody writing @agric and forestry On 11th??

I am

Whaylecs:
anyone writing in dept agric and foresty

yeah, I am.

Whaylecs:
i've seen the date guys

wow! Good for you. I have also seen mine, Room 204

Antoinne:
Seamione is actually right and you are wrong. Your sample space is incorrect. This is not a combination problem, but a permutation problem.

Selecting a multiple of 5 and an odd number from 1,2,4,5,6.
Will give 4x3x1 + 4x3x1.
To form a 3-digit number will give a space of 5x4x3
Probability = (2x4x3)/(5x4x3) = 2/5

Thanks bro, the case has long been settled though.

Mayour11:
pls,
2) A uniform beam is 12m long and has a mass of 50kg, and masses of 30kg and 40kg are suspended from its ends. At what point must the beam be supported so that it may rest horizontally?

6.5, if my assumptions are correct

bi0nics:
Add me on the whatsapp group..09025701919

I'm actually not the one to do that.

Cc: pr0ton

bi0nics:
Done..buh my biodata is empty..Help asap o.

That's how it is for everybody. So calm down

bi0nics:
BR0S..AYAF PAID...BUT IT STILL DY BRING THIS

clear the cookies or your phone or use another browser. You can also use a system if it persists

mathefaro:
Still very much available

ok, iyaff hear

Tolu91:
jiayu s3 for sale with all accessories. Goes for N33,000. Contact 08050899216

Is it the basic or advanced version and what's your location

sweetwealth:
The link is there at ur portal. U will see it wen u log in to ur post utme portal

ok Thanks

agentofchange1:
yea bro, i get, its ok now lets forget the whole thing... thanks anyways....

we are all learning.

You're welcome big boss

sweetwealth:
Thanks for the info, I just got my printed too

Please, how did you go about it? I have someone who has misplaced the one he printed

agentofchange1:
sorry check well, dia nine, i just made small mistake in writing d first group twice. den plus one other 1 i didn't include that will make it 10.

ur still repeating same sample space , no matter how far u go
hmm. i don't think i can explain this any further/longer maybe i should 'nt av used combinatorial approach for this question, i believe u will understand better some day, i really love ur type, who challenges/questions the 'lemmas' . I should have used the method u used ('wrongly ') , but i won't do that now as that will prompt another episode of arguments .m not really feeling strong. ..


nice solving.... keep the fire burning.

#shalom.

I don't know how to say this without sounding arrogant, but I strongly believe I perfectly understand what I need to in this context. It is you rather, who are refusing to change your perspective about the question, which is all good though.

Take a look at the question again If you don't mind

Three digit numbers re formed from the digits 1,2,4,5,6. If repetition is not allowed and a number is picked at random, find the probability that it's a multiple of 5 or an odd number.

The term "repetition is not allowed" here simply means when forming a 3-digit number (just one), no digit should appear more than once. So 124, 412, 214, 142, 421 and 241 are all in conformity with the rule guiding the formation of the numbers (i. 3-digit number from the digits 1,2,4,5,6 ii. No repetition of digits)
and are also totally different numbers such cannot be regarded as repeated numbers. Therefore, they all will make the sample space.
At this point if you still feel what about 3 of us here on this thread has said on this question is wrong, then I give up. Hoping that you change your perspective about it one day.

Mind you, it's not a matter of challenging/questioning your knowledge on this, but about not misleading lots of young ones who are gaining one or two things from this forum.

Happy weekend to y'all
#Peace

masperano:
Perhaps u should have added it up. So for instance 24+12=36 and then divided by 60 which will give you the required 3/5(a bit tricky).

Yeah, I think that what the text did. Actually, I also considered that but then I had to follow the rule of independent events.

To determine the probability of event X or event Y happening, when the two events are not mutually exclusive:
1.) Find the sum of the separate probabilities.
2.) From this sum subtract the probability that both events will occur.
P(X or Y) = P(X) + P(Y)-P(X and Y)

This will give us 12/60 + 24/60 - 12/60 = 2/5

agentofchange1:
hmmm,

i normally don't like arguing much, but let me try to explain this for the last time & if u guys think otherwise, that's Ok . so be it.

to explain this,

[b]lets say we have 5 students in a class ( named S1, S2, S3 , S4 & S5 ) then asked to form a group of three (probably, study group ) with no repetition allowed.(no repetition of same student(s) in the group or the sample space ) . already noted that S1 is the class rep. & S5 is the assistant class rep. , Now if a group is picked at random , whats the probability (or chance )of picking a group which has its last member as either S1 or S5 ? ( i. e either the class rep. or the assistant )

NB: In combination, we give attention to order of arrangements

total possible arrangement without repetition = 5C3
=10 ways

Now, possible sample space

S={(s1s2s4) ,(),(s1s2s5),(s1s2s6),(s1s4s5),(s1s5s6), (s2s4s5), (s6s4s5),(s6s4s1),(s2s6s5) . . . }

the possible groups continues to about ten of them , i just stopped there cuz, I've gotten what i needed , any more will result to repetition of already exiting group which goes against the given condition

Probability of having either s1 or s5 as last member of the group , could be (s1s2s5) or (s1s4s5) or (s2s4s5) or (s6s4s1) or (s2s6s5) or (s6s4s5)

= 1/10 + 1/10 + 1/10 +1/10 +1/10 + 1/10 =6/10 =3/5 [/b]
Now you guys be the judge

am off this question, happy solving .



Good night!!!!

Good morning bro. It's not that I find joy in arguing with you over this o as you be my oga patapata but I can't help but to point this out;
You said possible sample space

S={(s1s2s4) ,(s1s2s4),(s1s2s5),(s1s2s6),(s1s4s5),(s1s5s6), (s2s4s5), (s6s4s5),(s6s4s1),(s2s6s5) . . . }

Don't you think it's possible that the sample space could also possibly be

S={(s4s1s2) ,(s5s1s2),(s6s1s2),(s5s1s4),(s6s1s5), (s5s2s4), (s5s6s4),(s1s6s4),(s5s2s6), (s5s6s2) . . . } and in which your theory doesn't hold in this case.
Moreover, if you check your sample space, it's already up to 10 and you agreed that we could still continue, so many contradictions.
You see, my boss, whenever they use "no repetition" in this type of question, it simply means the same number can't show up twice in the same group like (s1s1s2). So, if you don't mind, you can also look this up.

Good morning to y'all
#Peace

jackpot:
that question is the same as ''probability that it is an odd number.'' probability= event occurence/sample space.


We have three boxes to be filled with the figures 1,2,4,5,6 without replacement. The last box only has 2 choices 1 and 5.
So, if the last box is 1, then we feed the other two boxes with 4 figures. No of ways this can be done is 4P2=12
Occurence= 4P2*2=24.
Sample space is 5P3=60

probability = 24/60=2/5.

Hmmmmmmmm. Someone should correct me, please.

You're absolutely correct and on point madam

chezholy:
I did.

Cool, that's what I also did

chezholy:
when I did mine, I got 24/60.
which is 2/5.

Did you consider both multiple of 5, odd number and the formula for independent events?

agentofchange1:
hmmm, i won't say much bro

let others shed more light.

Exactly, waiting for others to come in on this

agentofchange1:
c'mmon bro, the question says no repetition

5C3 means total number of ways we could form the 3-digit number without repetition

421 (124) ,521(125) , 621( 126) already exist in the sample space.

Oga General Ben, I think the "no repetition" in the question means we can't have numbers like 111, 222, 444, 555 and 666.
Moreover, 421 and 124 are not exactly the same number, they're just numbers with the same digits so we can't consider that as repetition because if that's the case, I don't think we should also have 124 and 125 because 1 and 2 have been repeated.
I hope you understand what I'm trying to say

agentofchange1:
total ways of selection = 5C3 = 10ways

sample space

S={(124),(125),(126),(245),(246),(145),(156),(645),(641),(265)}

check where we have an odd number or multiple of 5

thus, we get

6/10=3/5


hope you get?

No, I kinda disagree with your sample space, it can't just be 5C3. For instance, where Is 421, 521 and 621 in your sample space? At least, they're also of 3 digits which are odd numbers..

Good evening guys,
Please take a look at this question.

Three digit numbers re formed from the digits 1,2,4,5,6. If repetition is not allowed and a number is picked at random, find the probability that it's a multiple of 5 or an odd number.

From my calculations, I got 2/5 but the text says it's 3/5. Who is wrong?

tolaxin:
is C a pass

yeah, a credit pass