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In a game, one player throws two fair, six-sided die at the same time.
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22 Oct 2021, 06:56
22 Oct 2021, 06:56
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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. What is the probability that a player wins after playing the game once?
A. 1/3
B. 4/9
C. 5/9
D. 2/3
E. 3/4
A. 1/3
B. 4/9
C. 5/9
D. 2/3
E. 3/4
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Re: In a game, one player throws two fair, six-sided die at the same time.
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22 Oct 2021, 07:13
22 Oct 2021, 07:13
Carcass wrote:
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. What is the probability that a player wins after playing the game once?
A. 1/3
B. 4/9
C. 5/9
D. 2/3
E. 3/4
A. 1/3
B. 4/9
C. 5/9
D. 2/3
E. 3/4
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
= 5/9
Answer: C
Cheers,
Brent
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In a game, one player throws two fair, six-sided die at the same time.
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16 Oct 2022, 08:49
16 Oct 2022, 08:49
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
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
