I need help with my elementary statistics

**AlanY** · 07-17-2012, 01:06 PM

the probabilities of 5 individual events are;

1) .75
2) .75
3) .6
4) .85
5) .5

What is the probability for 1 out of 5; and 2 out of 5 etc?

many thanks in advance

**amleto** · 07-17-2012, 01:17 PM

do your own h/w. This is a badminton forum.

**Pizza Fish** · 07-17-2012, 01:58 PM

shouldn't this go in the chit-chat forum?
regardless, im not exactly sure what your math problem is....what do you mean by 1 of 5 or 2 of 5? the number 1 event or number 2 event or a combination? and if it's a combination, a combination of any two?

**AlanY** · 07-17-2012, 02:11 PM

what i wanted to work out is the probability of any 1 out of the 5 events occurring, or 2 events occurring etc.

for all 5 obviously is .75*.75*.6*.85*.5.
but what about just 1 or 2 of them?

**madbad** · 07-17-2012, 03:51 PM

Damn, it's a myth then... it appears not ALL Chinese are good at maths.

***jamesd20*** · 07-17-2012, 04:16 PM

The general principle i remember is and = multiply, or = add.

If it is two of any three it would be a combinatio, if any two, would be just add the two together.

Is it badminton related (I guess it is?)

***kwun*** · 07-17-2012, 04:28 PM

when all events have the same probability, then it is easy to do. but if they are not the same, then you have to find all the combinations of possibilities.

ie Pr(event1) * Pr(event2) * Pr(!event3) * Pr(!event4) * Pr(!event5) = 0.75 * 0.75 * (1-0.6) * (1-0.85) * (1-0.5).

that's one combination. you have to find all all combinations (there are 10 of them) and then just add them all together.

**AlanY** · 07-17-2012, 04:31 PM

Originally Posted by kwun

when all events have the same probability, then it is easy to do. but if they are not the same, then you have to find all the combinations of possibilities.

ie Pr(event1) * Pr(event2) * Pr(!event3) * Pr(!event4) * Pr(!event5) = 0.75 * 0.75 * (1-0.6) * (1-0.85) * (1-0.5).

that's one combination. you have to find all all combinations (there are 10 of them) and then just add them all together.

thanks, that's what i'm afraid of!

**madbad** · 07-17-2012, 04:33 PM

Originally Posted by jamesd20

The general principle i remember is and = multiply, or = add.

If it is two of any three it would be a combinatio, if any two, would be just add the two together.

Is it badminton related (I guess it is?)

Can't you guess? 5 events? Olympics around the corner... get it?

***kwun*** · 07-17-2012, 04:36 PM

similar analysis from a really old thread:

http://www.badmintoncentral.com/forums/showthread.php/15739-mathematical-reason-why-China-always-win-by-4-1.

***kwun*** · 07-17-2012, 04:40 PM

but you probably really want to find the probability of (*cough* hypothetically, of course *cough*) that at least 2 events occurs.

that means you only need to find:

1 - Pr(no event occur) - Pr(1 event occurs)

which is much more simple to do as there are only 6 combinations (1 for none + 5 for one)

***kwun*** · 07-17-2012, 04:42 PM

Originally Posted by AlanY

the probabilities of 5 individual events are;

1) .75
2) .75
3) .6
4) .85
5) .5

What is the probability for 1 out of 5; and 2 out of 5 etc?

many thanks in advance

Originally Posted by madbad

Can't you guess? 5 events? Olympics around the corner... get it?

i do think that 2) is a bit low, and 1) is a bit high!

**ctjcad** · 07-18-2012, 01:06 AM

Originally Posted by jamesd20

The general principle i remember is and = multiply, or = add.

If it is two of any three it would be a combinatio, if any two, would be just add the two together.

Is it badminton related (I guess it is?)

..you should've asked, "what is the probability it is badminton related?"..

**AlanY** · 07-18-2012, 01:25 AM

Ok, im trying to work out the chances of 1 badminton gold to the full house of ALL the countries at the OLYMPIC thus hoping for a short cut.
Oh well, its not going to be

**King's** · 07-24-2012, 10:00 AM

Originally Posted by madbad

Damn, it's a myth then... it appears not ALL Chinese are good at maths.

Perhaps up the ante by saying all Chinese Engineers are good at maths?

**uselessmail** · 07-25-2012, 03:52 AM

Could you clear some doubts here...

1. Are these events mutually independant events or absolutely random 5 events? In which case, you can simply multiply the individual events
2. What is the Sample Set? Are the numbers, probabilities of a certain event occurring all within a defined sample set? For eg. are the five numbers, the probabilities that China will win Gold in each of the five events. That means that there's a 75% probability that CHN gets gold in MS, 75% for gold in WS, 60% for Gold in MD.... an so on (hope you get the drift here!)

You need to define a Sample set and arrange the individual probabilities within so as to be able to calculate mixed probabilities. Then we can understand whether to simply multiply or use Kwun's extended formula!