Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59721

Jack has two dice, one has six equally probable sides, labeled 1, 2, 3
[#permalink]
Show Tags
08 Jul 2015, 03:45
Question Stats:
91% (01:07) correct 9% (02:04) wrong based on 85 sessions
HideShow timer Statistics
Jack has two dice, one has six equally probable sides, labeled 1, 2, 3, 4, 5, 6, and the other has seven equally probable sides, labeled 1, 2, 3, 4, 5, 6, 7. If Jack rolls both dice what is the probability that both of the numbers will be odd? A. 3/14 B. 2/7 C. 1/3 D. 1/2 E. 12/21 Kudos for a correct solution.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2977
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3
[#permalink]
Show Tags
08 Jul 2015, 03:56
Bunuel wrote: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3, 4, 5, 6, and the other has seven equally probable sides, labeled 1, 2, 3, 4, 5, 6, 7. If Jack rolls both dice what is the probability that both of the numbers will be odd?
A. 3/14 B. 2/7 C. 1/3 D. 1/2 E. 12/21
Kudos for a correct solution. METHOD1Probability that the number on first die is odd = 3/6 [Because 3 out of 6 faces are odd] Probability that the number on Second die is odd = 4/7 [Because 4 out of 7 faces are odd] Probability that Both Dice result in odd numbers = (3/6)*(4/7) = 2/7 Answer: Option B METHOD2Probability that the number on first die is odd = 3/6 [Because 3 out of 6 faces are odd] Probability that the number on Second die is odd = 4/7 [Because 4 out of 7 faces are odd] Probability that the number on first die is odd and second is even = (3/6)*(3/7) Probability that the number on Second die is odd and First is even = (3/6)*(4/7) Probability that the number on Both dice is even = (3/6)*(3/7) Favourable Probability = 1 Unfavourable Probabilityi.e. Favourable Probability = 1 [(3/6)*(3/7) + (3/6)*(4/7) + (3/6)*(3/7)] = 1  [(9+12+9)/42] = 1  [30/42] = 1  (5/7) = 2/7 Answer: Option B
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



CEO
Joined: 20 Mar 2014
Posts: 2560
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3
[#permalink]
Show Tags
08 Jul 2015, 04:27
Bunuel wrote: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3, 4, 5, 6, and the other has seven equally probable sides, labeled 1, 2, 3, 4, 5, 6, 7. If Jack rolls both dice what is the probability that both of the numbers will be odd?
A. 3/14 B. 2/7 C. 1/3 D. 1/2 E. 12/21
Kudos for a correct solution. Porbability = Total number of acceptable cases / Total number of all the cases In this case, total number of all the cases = 6*7 = 42 'Acceptable cases' = 1,1 1,3 1,5 1,7 Similarly with 3 and 5 for a total of = 3*4 = 12 Thus, the required probability = 12/42 = 2/7. Thus B is the correct answer.



Senior Manager
Joined: 28 Jun 2015
Posts: 279
Concentration: Finance
GPA: 3.5

Re: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3
[#permalink]
Show Tags
08 Jul 2015, 07:08
D1 = {1,2,3,4,5,6}; D2 = {1,2,3,4,5,6,7} Sample Space = 6*7 = 42. Odd Numbers: in D1 = {1,3,5}; in D2 = {1,3,5,7} Favourable Events = {1,1}, {1,3}, {1,5}, {1,7}, {3,1}, {3,3}, {3,5}, {3,7}, {5,1}, {5,3}, {5,5}, {5,7}, {7,1}, {7,3}, {7,5} = 12 events. Probability = 12/42 = 2/7. (B).
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15729
Location: United States (CA)

Re: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3
[#permalink]
Show Tags
08 Jul 2015, 21:13
Hi All, This question is a fairly standard probability question, so figuring out the individual probabilities and multiplying them together is a great approach for this prompt (and if you're not comfortable with that math yet, FireStorm's 'brute force' approach to list out all the possibilities is ALSO great). The answer choices are 'spaced out' enough that you can still get the correct answer with a bit of logic, but you do have to understand fractiontodecimals conversions and the basic math/logic behind rolling standard dice. Since a standard 6sided die has 3 odd 'sides' and 3 even 'sides', the probability of rolling an odd number on a single die is 1/2. By extension, the odds of rolling two odd numbers on two dice is (1/2)(1/2) = 1/4. Here though, the second die is a 7sided die and a little MORE than half of the numbers are odd (4/7 to be precise, but it's enough to say a 'little more than half'), so rolling a 6sided die and a 7sided die gives us a probability of getting two odds that is a little MORE than 1/4.... There's only one answer that's a little more than 1/4... Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



Math Expert
Joined: 02 Sep 2009
Posts: 59721

Re: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3
[#permalink]
Show Tags
13 Jul 2015, 03:46
Bunuel wrote: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3, 4, 5, 6, and the other has seven equally probable sides, labeled 1, 2, 3, 4, 5, 6, 7. If Jack rolls both dice what is the probability that both of the numbers will be odd?
A. 3/14 B. 2/7 C. 1/3 D. 1/2 E. 12/21
Kudos for a correct solution. 800score Official Solution:(B) For the first die there are 6 possible numbers, for the second die there are 7 possible numbers. When both are rolled together there are (6)(7) = 42 possible combinations. On the first die there are 3 odd numbers and on the second die there are 4 odd numbers. There are (3)(4) = 12 possible combinations where both numbers are odd. The probability that both numbers will be odd is 12/42 = 2/7.
_________________



Manager
Joined: 04 Apr 2010
Posts: 77
Schools: UCLA Anderson

Re: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3
[#permalink]
Show Tags
13 Jul 2015, 09:15
I think the 800Score official solution is too complex.
I'd just take the individual probabilities of rolling an odd number, and multiply them.
1/2 * 4/7 = 4/14 = 2/7
Answer: B




Re: Jack has two dice, one has six equally probable sides, labeled 1, 2, 3
[#permalink]
13 Jul 2015, 09:15






