Hello,
Can you please assist with this:
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?
OA: [spoiler]5/9[/spoiler]
Thanks a lot,
Sri
Probability of receiving a five or a one on either die,
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So, the player wins if he/she rolls AT LEAST one 5 or 1.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?
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
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Brent@GMATPrepNow wrote:So, the player wins if he/she rolls AT LEAST one 5 or 1.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?
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
Hello Brent,
Thank you very much for your excellent and detailed explanation.
Best Regards,
Sri
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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'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
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Hello Rich,[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
Thanks a lot for the alternate approach.
Best Regards,
Sri
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Brent@GMATPrepNow wrote:So, the player wins if he/she rolls AT LEAST one 5 or 1.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?
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
Hello Brent,
I was just wondering if we can apply the same technique of
P(Event A happening) = 1 - P(Event A not happening)
for the following as well or should this technique be strictly applied for only the ones that have "at least".
Mathematics, physics, and chemistry books are stored on a library shelf that can accommodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many mathematics books as physics books and the number of physics books is 4 greater than that of the chemistry books. Among all the books, 12 books are softcover and the remaining are hard-cover. If there are a total of 7 hard-cover books among the mathematics and physics books, what is the probability that a book selected at random is either a hard-cover book or a chemistry book?
OA: [spoiler]9/20[/spoiler]
m + p + c = 80/100(25)
=> m + p + c = 20
m = 2p and p = 4 + c
=> 2p + p + c = 20
=> 3p + c = 20
=> 3(4 + c) + c = 20
=> 12 + 3c + c = 20
=> 12 + 4c = 20
=> 4c = 8
=> c = 2
=> m + p = 18
Total Hard covers = 8
mathematics Hard cover + physics Hard cover = 7 => chemistry Hard cover = 1
Also, this means Total Soft Covers = 12
The probability that a book selected at random is either a hard-cover book or a chemistry book
= 1 - P (Not selecting a book that is either hard cover or chemistry)
However, I was not very sure how to solve from this point onwards.
Thanks a lot for your help.
Best Regards,
Sri
Brent,Brent@GMATPrepNow wrote:So, the player wins if he/she rolls AT LEAST one 5 or 1.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?
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
Please explain to me why it's not just P(A and B) = P(A) x P(B). You can have two outcomes on the first and two on the second which would be 2/6 * 2/6.
I clearly don't understand the principles.
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p(1,1) = 1/6 × 1/6 = 1/36
p(1,2) = 1/6 × 1/6 = 1/36
p(1,3) = 1/6 × 1/6 = 1/36
p(1,4) = 1/6 × 1/6 = 1/36
p(1,5) = 1/6 × 1/6 = 1/36
p(1,6) = 1/6 × 1/6 = 1/36
p(2,1) = 1/6 × 1/6 = 1/36
p(2,5) = 1/6 × 1/6 = 1/36
p(3,1) = 1/6 × 1/6 = 1/36
p(3,5) = 1/6 × 1/6 = 1/36
p(4,1) = 1/6 × 1/6 = 1/36
p(4,5) = 1/6 × 1/6 = 1/36
p(5,1) = 1/6 × 1/6 = 1/36
p(5,2) = 1/6 × 1/6 = 1/36
p(5,3) = 1/6 × 1/6 = 1/36
p(5,4) = 1/6 × 1/6 = 1/36
p(5,5) = 1/6 × 1/6 = 1/36
p(5,6) = 1/6 × 1/6 = 1/36
p(6,1) = 1/6 × 1/6 = 1/36
p(6,5) = 1/6 × 1/6 = 1/36
Therefore, the total sum is 20/36. So, there is more than two outcomes for the first and second, 2/6 does not account for all the possibilities.
p(1,2) = 1/6 × 1/6 = 1/36
p(1,3) = 1/6 × 1/6 = 1/36
p(1,4) = 1/6 × 1/6 = 1/36
p(1,5) = 1/6 × 1/6 = 1/36
p(1,6) = 1/6 × 1/6 = 1/36
p(2,1) = 1/6 × 1/6 = 1/36
p(2,5) = 1/6 × 1/6 = 1/36
p(3,1) = 1/6 × 1/6 = 1/36
p(3,5) = 1/6 × 1/6 = 1/36
p(4,1) = 1/6 × 1/6 = 1/36
p(4,5) = 1/6 × 1/6 = 1/36
p(5,1) = 1/6 × 1/6 = 1/36
p(5,2) = 1/6 × 1/6 = 1/36
p(5,3) = 1/6 × 1/6 = 1/36
p(5,4) = 1/6 × 1/6 = 1/36
p(5,5) = 1/6 × 1/6 = 1/36
p(5,6) = 1/6 × 1/6 = 1/36
p(6,1) = 1/6 × 1/6 = 1/36
p(6,5) = 1/6 × 1/6 = 1/36
Therefore, the total sum is 20/36. So, there is more than two outcomes for the first and second, 2/6 does not account for all the possibilities.
Last edited by talaangoshtari on Fri Jun 12, 2015 1:00 pm, edited 1 time in total.
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Hi e30sport,e30sport wrote:
Brent,
Please explain to me why it's not just P(A and B) = P(A) x P(B). You can have two outcomes on the first and two on the second which would be 2/6 * 2/6.
I clearly don't understand the principles.
Applying P(A and B) means finding the probability that we get 5 or 1 on the first die AND 5 or 1 on the second die.
To win the game we don't need to get 5 or 1 on BOTH dice.
Cheers,
Brent
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
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
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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 )