It was easier to me to figure out what the probability first of actually getting two consecutive numbers;
First the odds on any given roll of getting an even number is 3/6 or 1/2. The same is true of odd numbers. Because there is a 1/2 for even roll of getting an even, and a 1/2 change of getting and odd, we can think of this largely as a coin game. Now I must figure out how many possible combinations of even and odd there are in 4 rolls.
There are 5 cases; o=odd, e=even
1) All odd; oooo, the number of possible combinations = 4!/4! = 1
2)1 even the rest odd; eooo, the number of combinations with this is = 4!/(1!)(3!)= 4
3) 2 odd 2 even; ooee, number of combinations = 4!/(2!*2!) = 3*2*1 = 6
4)3 even 1 odd; eeeo, number of combinations = 4!/(1!3!)4= 4
5) All even; eeee, number of combinations = 4!/4! = 1
So, total there are 1+4+6+4+1=16 combinations. This could also be determine this by using the counting principle, which says that if you trying count the number of possible combinations, simply multiple the number of possiblities for each decision together. With this problem each roll will result in an even or odd number, so roll has 2 possibilities. So if there are 4 rolls;
2*2*2*2=16 (for four rolls)
Beacuse there are only 4 rolls and 2 possible results for each roll, it's easy to determine combinations where you would have 2 consecutive evens.
1) eeoo
2) oeeo
3) ooee
4) eeeo
5) oeee
6) eeee
These are the 6 ways in which you can roll consecutive even numbers. So there are 10 ways in which you can roll the dice and not get 2 consecutive even numbers.
10/16=.625, but this is not an answer choice.
There something missing from the question, or is my math above not correct?
Thanks,
Jared
Dice Roll
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Patrick_GMATFix
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Hey Shawshank. Is this an official question? Would you double check the answer choices please?Shawshank wrote:A dice is rolled 4 times. What is the of not getting 2 even numbers on 2 consecutive rolls.
a -0.25
b - 0.4
c - 0.5
d - 0.75
e - 0.8
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Rich@VeritasPrep
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As written above, this problem has some ambiguity issues. It doesn't specify whether we are looking for EXACTLY 2 consecutive even rolls or ANY 2 consecutive even rolls (i.e. a minimum of 2 consecutive even rolls).
But I'm assuming that it's asking about the probability of not getting 2 consecutive even numbers on ANY 2 consecutive rolls.
There are actually 8 ways out of 16 to get even numbers on any two consecutive rolls, 8 out of 16 ways not to:
No consecutive evens:
OOOO
OOOE
OOEO
OEOO
EOOO
EOEO
EOOE
OEOE
Consecutive evens:
EEOO
OEEO
OOEE
EEEO
EEOE
EOEE
OEEE
EEEE
If this is indeed what is meant by the problem, then we are interested in 8 possibilities out of 16, and the correct answer would be C.
But I would also be interested in hearing where this problem came from, and also whether or not it was indeed copied correctly.
But I'm assuming that it's asking about the probability of not getting 2 consecutive even numbers on ANY 2 consecutive rolls.
There are actually 8 ways out of 16 to get even numbers on any two consecutive rolls, 8 out of 16 ways not to:
No consecutive evens:
OOOO
OOOE
OOEO
OEOO
EOOO
EOEO
EOOE
OEOE
Consecutive evens:
EEOO
OEEO
OOEE
EEEO
EEOE
EOEE
OEEE
EEEE
If this is indeed what is meant by the problem, then we are interested in 8 possibilities out of 16, and the correct answer would be C.
But I would also be interested in hearing where this problem came from, and also whether or not it was indeed copied correctly.
Rich Zwelling
GMAT Instructor, Veritas Prep
GMAT Instructor, Veritas Prep
