In a game, one player throws two fair, six-sided die at the

at least based
so not getting 1 or 5 = 4/6
for two dice = 4/6 * 4/6 ; 4/9
1-4/9 = 5/9
IMO C

I feel the wording of the question could have been better.
Atleast a five or a one on either die- may be interpreted as 5,6 or 1
Better placement of the word five would be 'Roll a five or a one on at least either die'
VeritasKarishma Hope you all look into this!

hey buddy can you let me know why this does not work
prob(1 or 5 in first die) = 2/6=1/3
prob(1 or 5 in second die) = 1/3
So, required prob = 1/9

Hi AnirudhaS,

Your calculation is for the probability that BOTH dice show a '1' or a '5' - but the question asks for a '1' or '5' on EITHER of the dice. This means that there are 3 possible ways to win:

1) The 1st die matches and the 2nd does not: (1/3)(2/3) = 2/9
2) The 2nd die matches and the 1st does not: (1/3)(2/3) = 2/9
3) Both the 1st and 2nd die match: (1/3)(1/3) = 1/9

Total probability of getting at least one '1' or '5' = 2/9 + 2/9 + 1/9 = 5/9

You could also have solved this question by calculating the probability of NOT winning and subtracting that from 1. That would be...

1 - (2/3)(2/3) =
1 - 4/9 =
5/9

GMAT assassins aren't born, they're made,
Rich
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Given that In a game, one player throws two fair, six-sided die at the same time. If the player receives at least a five or a one on either die, that player wins. And we need to find What is the probability that a player wins after playing the game once?

As we are rolling two dice => Number of cases = \(6^2\) = 36

To win the player needs to get at least a five or a one on either die
=> Player can get 5 or 1 or 1, 5 on at least one die to win the game

So, possible cases are

Case 1: First roll get any number out of 1 or 5 in 2 ways
Second die get any number on the other die in 6 ways
=> 2*6 = 12 ways

Case 2: First roll get any number out of the six numbers except 1 or 5 as they are already considered in Case 1 => 4 ways
Second die get any number out of 1 or 5 in 2 ways
=> 4*2 = 8 ways

=> Total cases = 12 + 8 = 20

=> Probability that a player wins after playing the game once = \(\frac{20}{36}\) = \(\frac{5}{9}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems


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