probability strategy questions 1-3

[tkkishore]

What is the probability that the sum of two dice will yield a 4 or 6 ?

When you say there are 36 ways the 2 dice are thrown, then shouldnt you count (1,3), (2,2), (2,2), (3,1) as all possibilities for sum of 4 ? Therefore, i contest that the solution to problem 1 should be (4 + 6)/36 or 5/18. The 6 comes from 6 possible combinations that can produce sum of 6.

[StaceyKoprince]

Nope - that's a common error. When you calculate possible outcomes for probability, you always do one step at a time: first one die, then the other. Don't try to do these as a pair of dice rolled simultaneously. When there are multiple conditions, you always set one condition first, because that first condition then determines the possibilities for the second condition.

If you roll a 1 first, then you need the second to be a 3 or a 5. That equals two possibilities.
If you roll a 2 first, then you need the second to be a 2 or a 4. That equals two possibilities.
If you roll a 3 first, then you need the second to be a 1 or a 3. That equals two possibilities.
If you roll a 4 first, then you need the second to be a 2. That equals one possibility.
If you roll a 5 first, then you need the second to be a 1. That equals one possibility.
If you roll a 6 first, you're stuck - zero possibilities.

[Amit05]

What is the probability that the sum of two dice will yield a 4 or 6 ?

When you say there are 36 ways the 2 dice are thrown, then shouldnt you count (1,3), (2,2), (2,2), (3,1) as all possibilities for sum of 4 ? Therefore, i contest that the solution to problem 1 should be (4 + 6)/36 or 5/18. The 6 comes from 6 possible combinations that can produce sum of 6.

So, is the answer 2/9 ?

[shaji]

What is the probability that the sum of two dice will yield a 4 or 6 ?

When you say there are 36 ways the 2 dice are thrown, then shouldnt you count (1,3), (2,2), (2,2), (3,1) as all possibilities for sum of 4 ? Therefore, i contest that the solution to problem 1 should be (4 + 6)/36 or 5/18. The 6 comes from 6 possible combinations that can produce sum of 6.

So, is the answer 2/9 ?

Precisely!!!

[StaceyKoprince]

Yep, you've got it, guys!

[melissa_curran]

Hi, I don't understand why you look and rolling the dice simulataneously which would make the answer 5/18 on question 1 (incorrect). Can you elaborate on why you look at rolling one at a time?

[Ben Ku]

Hi, I don't understand why you look and rolling the dice simulataneously which would make the answer 5/18 on question 1 (incorrect). Can you elaborate on why you look at rolling one at a time?

Hi Melissa,

It's not really about rolling them one at a time; it's more about there being two dice. Imagine if one die is red and the other is white. Rolling a red 1 white 3 is different from rolling a red 3 and white 1.

Hope that makes sense.

[pellucide]

What is the probability that the sum of two dice will yield a 4 or 6 ?

When you say there are 36 ways the 2 dice are thrown, then shouldnt you count (1,3), (2,2), (2,2), (3,1) as all possibilities for sum of 4 ? Therefore, i contest that the solution to problem 1 should be (4 + 6)/36 or 5/18. The 6 comes from 6 possible combinations that can produce sum of 6.

I dont think you can count (2,2) twice. Also you cannot count (3,3) twice. So with the way you describe, the favourable outcomes should be
(1,3), (2,2), (3,1) - 3 outcomes for a 4-total
(1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes for a 6-total

So total favourable outcomes = 3+5
Total number of outcomes = 36
Probability = 8/36 = 2/9

[jnelson0612]

Correct, pellucide. Nice work!

[coreygillard]

Why can't you count (2,2) twice? You count (1,3) & (3,1)...isn't this the same principal? One die rolls a 2, the other rolls a 2. I'm confused why it works with 3 and 1, but not 2 & 2.

[tim]

imagine you have two different colored dice. rolling a blue 1 and a red 3 is different from rolling a red 1 and a blue 3. rolling a blue 2 and a red 2 is the same as rolling a red 2 and a blue 2..

[andre.renney]

I also had some difficulty understanding how two of the same numbers rolled twice are counted as one possibility and not two. However, I then listed out all the 36 possibilities and saw that the same number pairs (ex: 1-1, 2-2... 6-6) are only shown once.

Rolling a 2 for both pairs is only one possibility whereas rolling a 1 for Dice 1 and 3 for Dice 2 is not the same as rolling a 3 for Dice 1 and 1 for Dice 2.

[tim]

thanks!