sankp

IMO, Answer for Q-8: - C (median is 4) Too many links for Q-23. Not able to understand the how they are related;)

Let me attempt Q-34 IMO Ans: B Let x = incorrect paper record y = incorrect electric record Incorrect paper record also having incorrect elctric record (common set)=0.6x Incorrect electric record also having incorrect paper record (common set)=0.75y so, 0.6x=0.75y x=5/4 y Now, let T= total policy Common set=0.03T So, 0.03T=0.6x=0.75y x=T/20 y=T/25 So total number of Incorrect paper record and Incorrect electric record T/20+T/25-0.03T =3T/50 So total number of correct paper record and correct electric record =T-3T/50 = 47T/50 So, probability = 47/50=0.94

I did'nt see your SET TUTORIAL. ;) I guess it should be okay. In fact this set concept was quite clear to me. I messed up the calulation in hurry:yuck: Anyway back to the question.... Percentage of the survey participants liked more than one of the three products is 20 -[2*n (1n2n3)] = 20-10=10

D. n(1U2U3) = n(1) + n(2) + n(3) - n(1n2) - n(2n3) - n(1n3) + n (1n2n3) 85=50+30+20- [n(1n2) - n(2n3) - n(1n3)] +5 [n(1n2) - n(2n3) - n(1n3)] = 20

With due respect to araspee, I think araspee considered only 1 arrangement for white balls (-W-W-W-W-W-W-W-). But there are several other possible arrangements for white balls so that no 2 black balls are together. I am listing few of such possibilities.. -WW-W-W-W-W-W -WW-WW-W-W-W- -WW-WW-WW-W- ..... .... -WWW-W-W-W-W- -WWW-WW-W-W- -WWWW-W-W-W- If the question were " number of possible arrangements so that no 2 black balls and no 2 white balls are together", then araspee method would be correct.

Becuse from each nest you are collecting 2 eggs

IMO, (12!/(5!*7!))-(11!/(4!*7!)) = 462 (though it is not in answer choice):D

IMO it is C "Several of a certain bank’s top executives have recently been purchasing shares in their own bank." - At first, this is the evidence presented to counter the widely believed fact that the bank's financial condition is week. Then a counter evidence is presented - " However, the available information about the bank’s condition is from reliable and informed sources, and corporate executives do sometimes buy shares in their own company in a calculated attempt to calm worries about their company’s condition". Finally evidence and counter evidence are evaluated and concluded that "On balance, therefore, it is likely that the executives of the bank are following this example"

P( 1 couple together) = (3C1*8!*2!)/9! P( 2 couple together) = (3C2*7!*2!*2!)/9!

Total possible outcome 9C3 = 84 For Jaune X - at least 2 jars of yellow are required => Jaune X contains either 2 or 3 jars of yellow For Jaune Y - 1 jar of yellow is required So for any Jaune (X or Y), number of yellow jars could be 1 or 2 or 3 Now if 1 yellow jar is selected, other could be any 2 from balance 5. So possible combination is 4C1*5C2 =40 Similarly for 2 yellow jars, possible combination is 4C2*5C1 =30 For 3 yellow jars, combination is 4C3*5C0=4 Total possible combination of Jaune = 40+30+4 = 74 So, probability =74/84 = 37/42 HTH

Q-2 Possible outcome for at least 1 K= 4C1*8C1 + 4C2*8C0 = 38 Favourable outcome for 2 K = 4C2 =6 Probability for 2 K = 6/38 = 3/19

IMHO, Q-1 should read as King of Spade

My standarized approach will also give 84

If I understand the question correctly, No. of desired combination 12C3-(3C1*9C2+3C2*9C1+3C3*9C0)

Using Fermat's little theoram- 5^(13-1) =1|Mod 13|, 5 and 13 are coprime 5^12=1 |Mod 13| (5^12)^25 = 1^25 |Mod 13| 5^300 = 1 |Mod 13| 5^301 = 5 |Mod 13| So reminder is 5

Bullfighter, I think you are right . Answer should be 16/35 RRRB- 4/8*3/7*2/6*4/5 =2/35 RRBR =2/35 RBRR = 2/35 BRRR = 2/35 P(3R+1B) =8/35 Similarly P(3B+R) = 8/35 Total 16/35

To Bullfighter- All hats are identical. So for selection of 1 blue hat out of 4 blue hats is not 4C1. He can select any hat and any selction will yeild same result. Think this way.... A person can selcet 3 red hats and 1 blue hat but sequence of selection could be different e.g. First red, then red, then red and then blue. So there are 4 possible sequence -4!/3!. (3! is because 3 red hats are identical) RRRB RRBR RBRR BRRR So total 4 possible sequence. Similarly 4 possible sequences for 3 blue hats and 1 red hat. Hope I am correct!!

e) Total no of possibility = 8C4 =70 Exactly 3 red hats => 3 red hats + 1 blue hats => Total four hats (hats of same color are identical). So, possible sequence =4!/3! = 4 Similarly for exactly 3 blue hats, number of possible sequence =4 So total probability = (4+4)/70 = 4/35

A) 3c2*(1/3)^2*(2/3) + 3c3 * (1/3)^3 B) 3c1*(1/3)*(2/3)^2

C is correct 5 nos.- which are originally square of a number - 1,4,9,16,25. Therefore these numbers should be eliminated (Square root of square of these numbers will be perfect square). Now there will be total 30 numbers (25 as per Question stem and 5 additional to replace these discared numbers) S=summation of square of first 30 natural numbers = n(n+1)(2n+1)/6 = 30*31*61/6 = 9455 Summation of 1+16+81+256+625= 979 So, total = 9455-979 = 8476

IMO, Answer is C i found it through poe 77 is factor of 11! , but is not one bigger than a multiple of 6 385 is factor of 11! and is one bigger than a multiple of 6 463 is not a factor of 11! 1925 is factor of 11! , but is not one bigger than a multiple of 6

Neither sum nor mean- 4 individual numbers are 16 16,16,16,16,X

Median is $4300 64 percent of them have salary less than $5000 => 36% > $5000 60 percent is more than $4000 => 40% 5 percent is 4900, 11 percent is 4300 => 8% (24%-16%) remaining between $4000 and $5000 could be less than $4300 or geater than $4300. In both the cases median is $4300

What happens with n=19, x=9?

A= a^2*b*c N = (2+1)*2*3 = 12