2004 probability (1 Viewer)

[darlking]

Hi, can someone please help me with this question?
In a game, a turn involves rolling two dice, each with faces marked 0,1,2,3,4,5.

the score for each turn is calculated by multiplying the two numbers uppermost on the dice
what is the prob. that the sum of the scores in the 1st two terms is less than 45?

The solution does not make sense to me so hopefully someone can explain it in simple terms for a simple person

 

[coyney]

Is it 1-4/42 = 19/21

 

[darlking]

No, the answer is 1291/1296

 

[andybandy]

1 - (of sum >45) (draw a table)

 

[LivingLife]

Would someone be able to post the full worked solution to this? I also would like to know how to do this question

 

[andybandy]

<a href="http://www.codecogs.com/eqnedit.php?latex=1-P(\textup{Sum}\geq&space;45)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1-P(\textup{Sum}\geq&space;45)" title="1-P(\textup{Sum}\geq 45)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=P(25,25)&space;&plus;&space;P(20,25)&space;&plus;&space;P(25,20)&space;=&space;\frac{1}{36}.\frac{1}{36}&space;&plus;&space;\frac{2}{36}.\frac{1}{36}&space;&plus;&space;\frac{1}{36}.\frac{2}{36}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?P(25,25)&space;&plus;&space;P(20,25)&space;&plus;&space;P(25,20)&space;=&space;\frac{1}{36}.\frac{1}{36}&space;&plus;&space;\frac{2}{36}.\frac{1}{36}&space;&plus;&space;\frac{1}{36}.\frac{2}{36}" title="P(25,25) + P(20,25) + P(25,20) = \frac{1}{36}.\frac{1}{36} + \frac{2}{36}.\frac{1}{36} + \frac{1}{36}.\frac{2}{36}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=P(25,25)&space;&plus;&space;P(20,25)&space;&plus;&space;P(25,20)&space;=&space;\frac{5}{1296}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?P(25,25)&space;&plus;&space;P(20,25)&space;&plus;&space;P(25,20)&space;=&space;\frac{5}{1296}" title="P(25,25) + P(20,25) + P(25,20) = \frac{5}{1296}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=P(\leq&space;45)&space;=&space;1&space;-&space;\frac{5}{1296}&space;=&space;\frac{1291}{1296}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?P(\leq&space;45)&space;=&space;1&space;-&space;\frac{5}{1296}&space;=&space;\frac{1291}{1296}" title="P(\leq 45) = 1 - \frac{5}{1296} = \frac{1291}{1296}" /></a>

 

[darlking]

Can u plz explain it, i dont understand ur solution.

 

[andybandy]

Can u plz explain it, i dont understand ur solution.

okay, the question is asking, what is the probablity that the the sum of the first 2 scores ( each score is the mutiplication of the 2 dices) is greater than or equal to 45.
So youd find the probably of actually getting a 45 or greater, and then minus 1 from it, to get the probablity of not getting a 45 or greater?

 

[darlking]

Yes, i understand that, but when you use the 25 etc...????

 

[andybandy]

its the probably of rolling a score of 25? when the both dices land with 5 on the top

 

[bedpotato]

Yes, i understand that, but when you use the 25 etc...????

Draw a table for turn 1 and another for turn 2 (they're the same table but it's easier to visualise what's happening if you draw two). The score is calculated by adding a number from turn 1 and a number from turn 2. What numbers can you add that will be equal to or greater than 45? You can add 20 from the first turn to 25 from the second turn twice, or you can add 25 from both the first and second turns, or you can add 25 from the first turn to 20 from the second turn twice.

You're basically adding these numbers:

(20, 25), (20, 25), (25, 25), (25, 20), (25, 20)

P(sum equal to or greater than 45) = 2.P(20, 25) + P(25, 25) + 2.P(25, 20)

Now, the probability of obtaining a 20 is 1/36, and it's the same for 25

P(sum equal to or greater than 45) = 2(1/36 . 1/36) + (1/36 . 1/36) + 2(1/36 + 1/36)
= 5/1296

P(sum is less than 45) = 1 - P(sum equal to or greater than 45)
= 1 - 5/1296
= 1291/1296

lol hope that helps.