Probability Question (1 Viewer)

[bevstarrunner]

the way i did it...

P(anything) = (no. of favourable outcomes)/(no. of possible outcomes)
Favourable outcomes = (5,5 + 5,5), (5,5 + 5,4), (5,5 + 4,5), (5,4 + 5,5), (4,5 + 5,5) .'. there are 5 possibilities.
Possible outcomes = 36 x 36 = 1296

.'. P(>=45) = 5/1296
.'. P(<45) = 1291/1296

hope that helps

 

[JayWalker]

but its not out of 1296, there arent that many possibilities, as some of the numbers you cannot add the multiples together to get... i dont know, ill wait till a professioanl answer comes out

 

[bevstarrunner]

there are 36 possibilities from the first throw, right?
there are another 36 possibilities from the second throw.
.'. there are 36 x 36 possible outcomes

 

[JayWalker]

meh well what ever,, so i only get 116/120

 

[acmilan]

four rolls of the dice, 6 possibilities in each roll = 6^4 = 1296

 

[Sweet_Lemon]

wel...the answer is so obvious now....no point continuing argue~~ ekek let's debate something else...ekek

 

[JayWalker]

Again, im bored
Dum de dum

 

[idontgetit]

I get it now. think i could salvage a mark ?

 

[myg0t]

What are you trying to say?

The Answer is Here:
[>_ is greater or equal to]
P(<45) = 1 - P(>_45)
= 1 -[ P(Getting a 45) + P(Getting a 46) + ... +P(Getting a 50)
It is impossible to get a sum of 46, 47, 48 and 49,
HENCE
= 1 - [(5,5 + 4,5) + (5,5 + 5,4) + (5,5 + 5,5)
= 1 - [1/16 + 1/16 + 1/16]
= 1 - [3/16]
= 13/16
That is the answer that I got and my tutor agrees.. Dunno could be wrong.. general concensus on this board if that 1029 something answer...

::SHRUG::

How and/or why is it impossible?

 

[LozziPozzi]

hey for the first part, did every one get 1/6? because as soon as u get a zero, anything multiplied will give zero?

 

[idontgetit]

First part is 11/36, second is 1/9 and third (clenches teeth) is 1291/1296

 

[Sweet_Lemon]

not 1/6... 11/36 sound much more like d correct answer~~

 

[~ ReNcH ~]

I think the best way to solve this question conclusively is draw an entire table, which I painstakingly did in the exam yesterday.

0 1 2 3 4 5
---------------
0
1
2
3
4
5

Multiply them, complete the table and you find that there are 36 possible outcomes, of which 2 are "20" and 1 is "25".
.'. P(20) = 2/36 = 1/18 & P(25) = 1/36

.'. P(sum >= 45) = P(20+25) + P(25+20) + P(25+25)
= 1/18x1/36 + 1/36x1/18 + 1/36+1/36
= 5/1296
.'. P(sum < 45) = 1291/1296

 

[bevstarrunner]

First part is 11/36, second is 1/9 and third (clenches teeth) is 1291/1296

100% correct

 

[icebreaka]

P(<45) = 1 - [ (1/36 * 1/36) + (1/36 * 1/36) + (1/36 * 1/36) + (1/36*1/36) + (1/36*1/36) ]

= 1 - ( 5/1296)
= 1291 / 1296

Final Answer: 1291 / 1296 :d

 

[A2RAYA]

thats what i got two but i don't think it was right
it looked wrong

well i guess i was way off.....it was sumthing outta 72...hmmm...and i thought that was an easy question....never again

 

[Smurfwow]

ok time to settle this once and for all

I wrote a program which does the two turn thing a billion times (ill put the code at the end of this post if you're interested)

the ratio of sums less than 45 was 0.996138238, which very much looks like it converges on 0.996141975 (which is 1291/1296 to 9dp).

so unfortunately for me and some others here, the answer is 1291/1296.

Random rand = new Random();
Random extraRandom = new Random();

public void run()
{
	int lessThan45 = 0, theOtherOne = 0;
	int score1 = 0, score2 = 0;
	for(int i = 0; i < 1000000000; i++)
	{
		score1 = haveTurn();
		score2 = haveTurn();
	
		if(score1 + score2 < 45) 
			lessThan45++;  
		else 
			theOtherOne++;
	
		if(i%10000000==0)
		{
			rand = new Random(extraRandom.Next());
		}
	}
	double ratio = lessThan45 / 1000000000.0d;
	Console.WriteLine(ratio);
}

public int haveTurn()
{
	return rollDice() * rollDice();
}

public int rollDice()
{
	return rand.Next(0, 6); // this gives us 0 through 5
}