Brian plays a game in which two fair, six-sided dice are rol

Die is singular of dice.

Hi Bunuel
Could you clarify, "Brian will win in cases when both dice show an even number as well as in cases when an odd number on one die is less than an even number on another."

The question stem does not say anything about the odd number has to be less than the even number. IMO, the winning combinations are:

1.) e-e: 3/6*3/6
2) o-e
1,2:1,4:1:6
3,2: 3,4: 3,6
5,2: 5,4: 5,6

Winning prb: 3/6*3/6+ (9/36)*2= 1/4+ 1/2= 3/4

Thanks

For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money.

So, for example:
In case of (2=even, 4=even) he wins 2 + 4 = $6.
In case of (2=even, 5=odd) he losses 2 - 5 = $3.
In case of (6=even, 1=odd) he wins 6 - 1 = $5.

Hope it's clear.

Thanks make sense. Need to read the stem more carefully.

This is an interesting question and needs you to read the question stem attentively.
We are sure that Brian will win in the case when there is an even number and loose the same amount if there is an odd number.

Bifurcating the cases.

1. Brian will win some money if there are even numbers on both the dices.
2. Brian will win some money if the even number on one die is more than the odd number. Here Brian will win the money equal to the difference of the two numbers on the dices.

Case 1:
Even numbers on both dices.
Since there are 3 even numbers on a dice: 2, 4, 6
This is possible in 3*3 = 9 ways

Case 2:
Even number is greater the odd number
This is possible in the following ways
(2, 1), (4, 1),(4, 3), (6, 1), (6, 3), (6, 5)
These numbers can be interchanged on both the dices.
Hence a total of 12 cases.

Total outcomes = 6*6 = 36
Probability = 12+9/36 = 21/36 = 7/12

Option B.

VERITAS PREP OFFICIAL SOLUTION:


First, consider how Brian can win or lose: two even numbers guarantee a win, while two odd numbers guarantee a loss. If he gets one odd and one even, he wins money if the even is greater than the odd and loses money if the odd is greater than the even. If he rolls a 6, he can roll any odd number and still make money. If he rolls a 4, then odd must be 1 or 3, and if he rolls a 2, then the odd must be 1. The probability of making money is then:

Even AND even: 3/6 * 3/6 = 9/36 = 1/4

One even, one odd:

6 AND any odd: 1/6 * 3/6 = 3/36

Any odd AND 6: 3/6 * 1/6 = 3/36

4 AND 1 or 3: 1/6 * 2/6 = 2/36

1 or 3 AND 4: 2/6 * 1/6 = 2/36

2 AND 1: 1/6 * 1/6 = 1/36

1 AND 2: 1/6 * 1/6 = 1/36

Total = 12/36 = 1/3

The probability of making money is then 1/4 + 1/3, or 7/12.

If you are short on time here, its not that hard to guess.

The odd numbers on a die sum up to 9 (1+3+5) but the even numbers sum up to 12 (2+4+6).

Logically the correct answer will be a little more than 50 %. Only B fits.