Math Help !!!

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I3east

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Okay, i have been set some maths homework and i cant do it at all.

I have to find the probability of scoring more than one. Here are the results;

1 - 178
2 - 168
3 - 175
4 - 179

What i don't get is do you add them all up or what ? Help would be appreciated :)

EDIT: Is this right ?

Is the relative frequency of 3 - 0.04 when this is the result

3 - 76
 
Last edited:
Same here :P

---------- Post added at 08:39 PM ---------- Previous post was at 08:35 PM ----------

It's on MyMaths and i need over 70% to complete it and i can only get 58% Argh !!!
 
I'm assuming that there are only 4 possible outcomes of scoring which are 1,2,3 and 4. I am also assuming that the sample is representative of the population. Correct me if I am wrong.

Add up the total number of results. This will be the denominator. Add up the results when X>1. This will be the numerator.

178 + 168 + 175 + 179 = 700.

P(X>1) = (168 + 175 + 179) / 700

= 522/700

= 261/350 or 0.7457

Either it's that simple or I have misunderstood the question!
 
I3east said:
Okay, i have been set some maths homework and i cant do it at all.

I have to find the probability of scoring more than one. Here are the results;

1 - 178
2 - 168
3 - 175
4 - 179

What i don't get is do you add them all up or what ? Help would be appreciated :)

EDIT: Is this right ?

Is the relative frequency of 3 - 0.04 when this is the result

3 - 76
Click to expand...

the results ?- Is it the numbers of matches that have ended with 1 goals scored, 2 goals scored etc.
 
Nixon said:
I'm assuming that there are only 4 possible outcomes of scoring which are 1,2,3 and 4. I am also assuming that the sample is representative of the population. Correct me if I am wrong.

Add up the total number of results. This will be the denominator. Add up the results when X>1. This will be the numerator.

178 + 168 + 175 + 179 = 700.

P(X>1) = (168 + 175 + 179) / 700

= 522/700

= 261/350 or 0.7457

Either it's that simple or I have misunderstood the question!
Click to expand...

That would be 75% so seems logical looking at the numbers. ;)
 
It is the amount of times that a triangular based dice landed on 1-4

---------- Post added at 08:53 PM ---------- Previous post was at 08:52 PM ----------

Thanks Nixon, i tried that method but it hasn't worked, may just not do it .
 
I3east said:
It is the amount of times that a triangular based dice landed on 1-4

---------- Post added at 08:53 PM ---------- Previous post was at 08:52 PM ----------

Thanks Nixon, i tried that method but it hasn't worked, may just not do it .
Click to expand...

Is the die fair? And how many sides does it have? 4?

If it is fair and has 4 sides, then the probability that X>1 is exactly 75%.

If it isn't fair then we have to use the sample data, however this is only an estimated probability.
 
Yeah it is a fair dice, it doesn't matter people, i have got 100% and 81% on my other two ones he set me so i will be ok, thanks for you help :)
 
Nixon hit the nail on the head.

I got 0.126 when 3=76.

You're looking at the probability that you get a certain number from all the throws. 3 is seen 76 times in 601 throws of the die, so the probability of seeing 3 is 76/601. Think of it as a simple coin toss. If you throw it twice, and see tails once, the probability is 1/2.

---------- Post added at 09:14 PM ---------- Previous post was at 09:07 PM ----------

Also, if you're struggling with MyMaths homework just let someone else log on and do it for you. ;)
 
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