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How does toss being consecutive adds anything to the question ? What is the difference between these two questions : A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses? vs A coin is tossed 6 times. What's the probability of getting at least 4 heads ? ...

by beat_gmat_09

Sat Dec 25, 2010 8:03 pm
Forum: Problem Solving
Topic: Probability
Replies: 17
Views: 3021

The question asks for at least 4 heads on consecutive tosses . 1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15 This will contain some ways such as (HTHTHH, THHHTH, etc...) which doesn't satisfy the criteria. Correct me if I ...

by beat_gmat_09

Sat Dec 25, 2010 7:49 pm
Forum: Problem Solving
Topic: Probability
Replies: 17
Views: 3021

Thanks for the above But I think the two probabilities should be different. I mean, "what percent of all the possible subcommittees that include Michael also include Anthony"----Here we are calculating the % with the base as "no. of teams having Michael"..Hence the answer would ...

by beat_gmat_09

Sat Dec 25, 2010 7:37 pm
Forum: Problem Solving
Topic: Manhattan cat1 ---Probability --committee of three ppl
Replies: 4
Views: 1384

P = 4Q , Q = 4R , R = 4S , S Bring each term in terms of S. P = 4*4*4S , Q = 4*4S , R = 4S , S 64S , 16S , 4S , S The ratio 13:4 will come when Q&R are chosen (16S+4S = 20S) and P&S are chosen. Ratio P&S/Q&R = 65S/20S which is equal to 13/4 Comparing time taken by P&S and Q&R...

by beat_gmat_09

Sat Dec 25, 2010 12:55 am
Forum: Problem Solving
Topic: Combined Work.
Replies: 6
Views: 1958

Hi, The question asks for at least 4 heads. 1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15 2nd we get 5 heads remaining 1 has to be tail. This can happen - HHHHHT, number of ways heads can occur = 6!/5! = 6 3rd we get all 6 ...

by beat_gmat_09

Fri Dec 24, 2010 11:12 pm
Forum: Problem Solving
Topic: Probability
Replies: 17
Views: 3021

Thought i'd share this- Consider 2 examples , first with 3 persons, second with 6. Take for example there are 3 persons EMX, and E should be behind M. Let's see how can this be arranged. E _ _ _ E _ Now 1st case - M can occupy 2 places E M X or E X M, hence number of ways = 2 2nd case - M can occupy...

by beat_gmat_09

Fri Dec 24, 2010 9:40 pm
Forum: Problem Solving
Topic: A,A,B,B,B,C,D,E arrangement question
Replies: 6
Views: 4350

You can't do that, here is why : 6 > 5 7 > 5 Can I do 6/7 > 5/5 or 6/7 > 1 No, Right ? Because in an inequality we don't know by how much LHS can be greater/smaller than RHS. OR 7/6 > 5/5 7/6 > 1 (Here we are just lucky, that the inequality still holds). :) Picking numbers was my last resort, actua...

by beat_gmat_09

Thu Dec 23, 2010 9:26 pm
Forum: Data Sufficiency
Topic: x,y and z
Replies: 4
Views: 1302

x,y and z

This if from OG If x,y, and z are positive numbers, is x > y > z ? 1) xz > yz 2) yx > yz x,y and z are positive integers(numbers) and we have to determine if x > y and y > z From 1) : as z is positive the inequality can be divided by z on both sides, thus x > y Not sufficient. From 2) : again as y i...

by beat_gmat_09

Thu Dec 23, 2010 7:18 pm
Forum: Data Sufficiency
Topic: x,y and z
Replies: 4
Views: 1302

Following can be the only possible combinations: (3337), (3346), (3355), (3445) and (4444) So total number of ways of distributed (3337) among 4 is = 4 Number of ways of distributing (3346) is = 12 Number of ways of distributing (3355) is = 6 Number of ways of distributing (3445) is = 12 Number of ...

by beat_gmat_09

Thu Dec 23, 2010 2:47 am
Forum: Problem Solving
Topic: Orange
Replies: 4
Views: 1253

Orange

If 16 oranges are distributed among 4 children such that each gets at least 3 oranges, the number of ways of distributing them is

A) 30
B) 210
C) 15
D) 35
E) 40

I don't have the OA for this.

by beat_gmat_09

Thu Dec 23, 2010 1:50 am
Forum: Problem Solving
Topic: Orange
Replies: 4
Views: 1253

C(5,3)*3! = C(5,2)*6 = 10*6 = 60

by beat_gmat_09

Thu Dec 23, 2010 1:45 am
Forum: Problem Solving
Topic: Combination
Replies: 5
Views: 1143

Total students = 150
Honor failed = 115
Honor made = 150 - 115 = 35
Honor Made = 10 + Varsity , Varsity = 35 - 10 = 25
Both = 10
Total = H + V - Both + Neither
Neither = Total - H - V + Both
= 150 - 25 - 35 + 10
= 150 - 50
= 100

Probability = Neither/Total = 100/150 = 2/3

by beat_gmat_09

Tue Dec 21, 2010 8:36 pm
Forum: Problem Solving
Topic: Kaplan premier--Honor & sports
Replies: 4
Views: 1487

Hi, The word - REVIEW contains 6 alphabets with 2 E's Tie these E's together - RVIWEE consider EE as 1 entity total entities = 5, number of ways to arrange these = 5! = 120 The order of E's doesn't matter here as they are same - indistinguishable. Total number of ways to arrange 6 alphabets with 2 E...

by beat_gmat_09

Tue Dec 21, 2010 8:16 pm
Forum: Problem Solving
Topic: Forbidden arrangement question
Replies: 4
Views: 1541

Thanks for the effort Rahul, but i am not comfortable with plotting graphs :(

by beat_gmat_09

Mon Dec 20, 2010 7:19 am
Forum: Data Sufficiency
Topic: x and y
Replies: 4
Views: 1133

Thanks for the link Rahul.
I feel the algebraic approach time consuming sometimes, it works for certain problems.
I was look for picking numbers (proper numbers) for this problem to evaluate the question.
I think it works faster for some problems.

by beat_gmat_09

Mon Dec 20, 2010 4:22 am
Forum: Data Sufficiency
Topic: x and y
Replies: 4
Views: 1133