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[roosterman]

Hey,
can anyone help me with this Qs. its prol simple but i find it a bit tricky


1. a box contains 3 black pens, 4 red pens and 2 green pens. if i draw out 2 pens, find the probabilitly that they are both red.


for this question is it- 4/9 times 4/9??????

or 4/9 times 3/8. thanx

 

[2 Dea]

It depends on whether they are put back in after you draw the first one out. But I'm guessing that it means one after the other, so its 4/9 * 3/8.

 

[roosterman]

thanx

also 4 this Q. for what values of x is the curve y= 9 - x^2 concave downwards?

is it all real x??

 

[SoulSearcher]

let f(x) = 9-x^2

f'(x) = -2x

f''(x) = -2

for concave down, f''(x) < 0

since -2 is smaller than 0, the curve is concave down for all real x

at least i think thats how they do it

 

[pLuvia]

y=9-x²
y'=-2x
y"=-2

Since y"<0 the curve is concave down hence for all real x

 

[Riviet]

for this question is it- 4/9 times 4/9??????

If the balls are replaced each time, then it is 4/9 x 4/9

 

[roosterman]

what about if they are drawn simultaneously????

 

[webby234]

what about if they are drawn simultaneously????

That's the same as if it was not replaced after the first draw.

 

[Riviet]

If the question uses "and" you will usually need to multiply the probabilities of each occuring.
If the question uses "or" you will usually need to add the probabilities of each occuring.

 

[webby234]

And if the question uses "at least" it is usually easier to find the probability of it not happening and subtract from one.

 

[roosterman]

hey i just did my maths exam on probability.
im such a tool i think changed the corrrect answer.
it was ....... Two golfers are estimated to have the probabilty of 1/3 and 1/4 respectively of winning a tournament. if they play in a tournment where it is possible to come equal first, find the probability that:

a) one wins

what do u guys think the answer is????

 

[Riviet]

hey i just did my maths exam on probability.
im such a tool i think changed the corrrect answer.
it was ....... Two golfers are estimated to have the probabilty of 1/3 and 1/4 respectively of winning a tournament. if they play in a tournment where it is possible to come equal first, find the probability that:

a) one wins

what do u guys think the answer is????

This is how i would do it.
I would list all the possibilites that could occur and let them all add to 1 since total of possibilities always equals 1.
Here are the possibilities:
* They are equal first, implying golfer A and golfer B win =1/3x1/4=1/12
* Either one of them wins, implying golfer A or golfer B wins =1/3+1/4=7/12
* Neither win=1-P(at least 1 golfer winning)=1-8/12=1/3

After i think about it, i think it's still 7/12, unless the question also includes them both winning, then you would just add the first two proabilities to get 3/4.
Someone please confirm me on this as i'm not 100% certain.

 

[insert-username]

I worked it out like this:

If the question wants the probability of only one winning, then it is the probability of one winning the tournament plus the probability of the other winning the tournament, minus the chance that they will come equal first. That is, 1/4 + 1/3 - 1/12 = 1/2.

If the question wants the probabiity of one winning, including an equal first, then it's simply the two probabilities added together - 7/12: 1/4 + 1/3.


I'd say the question wants the first answer, the P of just one winning.

I_F

 

[roosterman]

the answer

is 1) (A wins x B loses) = (1/3 x 3/4) +
2) (A loses x b wins) = (2/3 x 1/4)

=5/12

i had it the 1st time then i changed it!

 

[Riviet]

Oh... i see. I knew i had something wrong with mine.

 

[insert-username]

You shouldn't have to ask the same question in two different threads. For 1a, the answer is 336 ways, 8P3.


I_F