Brent@GMATPrepNow wrote:

gmattesttaker2 wrote: In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?

So, the player wins if he/she rolls AT LEAST one 5 or 1.
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
= 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
= 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
= 1 - [ 4/6 x 4/6]
= 1 - [16/36]
= 20/36
= [spoiler]5/9[/spoiler]

Cheers,
Brent

[email protected] wrote:Hi Sri,

Brent's solution is great (and it's exactly how I would approach the question). There is another way to approach it though. Since you're rolling 2 dice, there aren't that many possible outcomes (just 36 in total), so you COULD use brute force and just write them all down:

We're looking for the number of outcomes that include AT LEAST a 1 or a 5.

1,1
1,2
1,3
1,4
1,5
1,6

2,1
2,5

3,1
3,5

4,1
4,5

5,1
5,2
5,3
5,4
5,5
5,6

6,1
6,5

Total possibilities = 20

Probability of rolling at least a 1 or a 5 on two dice: 20/36 = 5/9

GMAT assassins aren't born, they're made,
Rich

Brent@GMATPrepNow wrote:

gmattesttaker2 wrote: In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?

So, the player wins if he/she rolls AT LEAST one 5 or 1.
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
= 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
= 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
= 1 - [ 4/6 x 4/6]
= 1 - [16/36]
= 20/36
= [spoiler]5/9[/spoiler]

Cheers,
Brent

Brent@GMATPrepNow wrote:

gmattesttaker2 wrote: In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?

So, the player wins if he/she rolls AT LEAST one 5 or 1.
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
= 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
= 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
= 1 - [ 4/6 x 4/6]
= 1 - [16/36]
= 20/36
= [spoiler]5/9[/spoiler]

Cheers,
Brent

Anantjit wrote:It can be like below :-
Probability of 5 or 1 on either dice= P(5) or P(1) on 1st dice OR P(5) or P(1) on 2nd Dice
i.e P[A]+P-P[A and B]
P[A]=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
P=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
P[A and B]= 1/3+1/3-[1/3*1/3]=2/3-1/9=5/9

This is the right idea, but you want to minimize the amount of clutter on your page during the test, or it's easy to confuse yourself with too much superfluous jargon. (It also probably wouldn't help someone who doesn't understand the idea in the first place, as it's a bit of a wall of text: keeping it simple and clean is good for the teacher and the student )