Austin_TX

7C2 * (1/2)^7 = 21/128

Congratulations Ursula ! This is awesome. I would really appreciate if you can share: - what kind of problem you got fom probability/permutation/combination/standard deviation. Were those really difficult or at the same level we usually see at this forum. - did questions got easier after you 'nailed' the first few difficult questions. - also, did u ran out of time in any section? if yes did you left a few questions unanswered or guessed. Thanks.

I don't uderstand the answer to sceanrio number (1). It should not matter how small and large a set is. It is the range that should matter. I think, a set of only two numbers {1, 1000} will have higher standard deviation than a set of 10 elements {5,5,5,5,5,5,5,5,5,5}.

This question was discussed earlier on this forum, but no answer was provided by the person who posted this question. There was lot of discussion but no consensus. So here it is again: A club has a membership of 9 men and 12 women...21 people total. A 5-member committee is to be selected to attend a conference. How many committees with no more than three women be formed?

I think this question was discussed earlier in this topic, but there was no answer provided and there was lot of discusson but no agreement on answer or explanation. Here it goes again: John has 7 bananas and 3 kiwis. He needs to divide the 10 in two parcels, so there is an equal total number of fruit in either parcel and so that there is at least one kiwi in each parcel. a) 21 b) 35 c)60 d) 105 e) 120 I calculated the result to be 105. Do you guys agree? I just want to make sure if my approach and answer is correct.

Hi vknittala: I follow your logic but I don't understand when you state "5 => means selecting the 100th place from the remaining 5 numbers in 5c1". For 100th place there are 7 numbers available, why you are saying only 5 numbers available. Can you please explain. Thanks.

I think this question was discussed earlier in this topic, but there was no answer provided and there was lot of discusson but no agreement on answer or explanation. So here it is again: From the word ZEBRA how many different letter combinations can be arranged by using the foregoing letters? a) 20 b) 60 c)100 d) 120 e) 150

87 * .0000125 * x = .018 x = 16 and some fraction but if only full servings are allowed x = at least 17 servings

Is answer 17

Your answer is correct, and you explained it very well. Thanks.

Your answer is correct, and you explained it very well. Thanks.

Thirteen cards numbered through 1 through 13 are placed in an urn. If five cards are drawn at random, what is the probability that the numbers are sequential? The drawn cards may be rearranged if necessary. Answer = 1/143

If you roll 3 dices, what is the probability the numbers will add up to 6? Answer = 5/108

Is answer B ?

Is answer E ? I don't know if this is correct, but here is my analysis: Let the big countertop has side = b; area of countertop = b^2 Let the tile inlay has side = a; area of tile = a^2 The strip has width = x; which means b=a+2x Area of untiled strip= area of countertop – area of tile inlay = (b^2) – (a^2) (a^2) / ((b^2) – (a^2)) = 25/39 (a^2) / (b^2) = 25/64 a/b=5/8 we know that b= a+2x a / (a+2x) = 5/8 a = (10x)/3 Plugging in the values from answer choices x = 3/2; a = 5; b = 8; a/b = 5/8; (a^2) / ((b^2) – (a^2)) = 25/39 x = 3; a = 10; b = 16; a/b = 5/8; (a^2) / ((b^2) – (a^2)) = 25/39 x = 9/2; a = 15; b = 24; a/b = 5/8; (a^2) / ((b^2) – (a^2)) = 25/39 All of them are valid.

That explains it. Thanks.

Hi DaffyDuck, The answers are: 1) 384 2) 10 3) 90 Can you please explain your approach. Thanks.

1) Four married couples bought eight seats in a row for a football game. In how many ways can they be seated if each couple is to sit together? Ans=384 2) A carton of eggs contain two bad ones. In how many ways can we select three of the 12 eggs so that both of the bad ones are included? Ans= 10 3) A carton of eggs contain two bad ones. In how many ways can we select three of the 12 eggs so that one of the bad ones is included? Ans=90

(Prob of male)(Prob of Men) + (Prob of Female)(Prob of Women) = (5/11)(8/11) + (6/11)(3/11) = 58/121

I was not clear about the question I assume by 3 + 5/root(5-x), you meant (3+5)/root(5-x) So, (3+5) / root(5-x) = 6 8 / root(5-x) = 6 8 = 6(root(5-x)) squaring both sides 64 = 36(5-x) 64 = 180 – 36x x= 3.2

Hi Truman, You cannot add (2^2x) + (2^2x) = 4^2x. This is mathematically wrong. If you do that, then it means you are saying: (2^3) + (2^3) = 4^3, which is wrong. This is where you are making mistake. Please see my explanation for correct approach.

The correct one: (2^2x) + (2^2x) = (2^100) Let A = (2 ^2x) A + A = (2^100) 2A = (2^100) A = (2^100) / 2 A = (2^99) Since A = (2 ^2x) (2 ^2x) = (2 ^99) If base is same, exponents are equal to each other 2x = 99 x = 99/2

I would really appreciate if someone can please explain the concept/technique of cyclicity. Thanks.

The answer should be 99/2

It is simple. The formula for handshakes among n people is = n(n-1)/2 The logic for the formula is that n people will shake hands with n-1 people. (Because a person won't shake hands with himself, therefore n-1 is used) So the total no. of handshakes is n(n-1). BUT we just double counted the hand shakes, because we counted that person A shake hands with person B and ALSO counted person B shaking hands with A. We have to correct for this double counting by didving by 2. Therfore the number of handshakes among n people = n(n-1)/2 I used this formula twice to came up with the answer. There are total of 18 reps Total no. of handshakes among all reps (including own company) = (18 X 17)/2 = 153 No. of handshakes among one company’s own reps = (3 X 2)/2 = 3 No. of handshakes among 6 company’s own reps = 3 X 6 = 18 Total no. of handshakes among all reps excluding own company’s reps = 153-18 = 135