In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47 b. (13^4) x 27 x 47 c. 48C6 d. 13^4 e. (13^4) x 48C6
I am lost.. Got several methods and several different answers.. bhavin I think u r d one to rescue.. :
Possible answers: 1)13C1 x 13C1 x 13C1 x 13C1 x 48C2 (not in option) 2)13C1 x 13C1 x 13C1 x 13C1 x 48 x 47 3) 1 x 1 x 1 x 1 x 48C2
FYI...I saw this question in IMS material today. There is a typo in answer choice B. It needs to be 24 instead of 27. Which translates to 13^4 * 48C2 as the answer.
FYI:I don't have IMS material.. :) neways to update u even 13^4 * 48C2 is not d answer if at all IMS ppl have mentioned dat as d OA..
for the correct answer refer to bhavin's post..
Vikram2010 Says
FYI...I saw this question in IMS material today. There is a typo in answer choice B. It needs to be 24 instead of 27. Which translates to 13^4 * 48C2 as the answer.
If x is equal to the sum of even integers from 400 to 600, both inclusive, and y is the number of even integers 400 to 600, both inclusive, what is the value of x+y ?
If x is equal to the sum of even integers from 400 to 600, both inclusive, and y is the number of even integers 400 to 600, both inclusive, what is the value of x+y ?
A)50601 B)50600 C)56002 D)57002 E)50000
Remember a simple formula: when the end terms are inclusive{divided by 2 because of even integers, a more generalized form is {(n2-n1)/N}+1 where N is the dividing number)
no. of terms={(n2-n1)/2}+1
if non-inclusive no. of terms={(n2-n1)/2}
here we have y={(600-400)/2}+1=101
Now Sum of even terms
Use a simple Sum of AP=n(a+l)/2 where a-first term and l-last term of the series.
A certain Junior class has 1000 students, and a certain senor class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A certain Junior class has 1000 students, and a certain senor class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A) 3/40000 B) 1/3600 C) 9/2000 D) 1/60 E) 1/15
60 sibling pair means 60 students in Junior Class and 60 students in senior class.
P(e)=(60C1/1000C1)*(60C1/800C1)=9/2000 i.e. option C
If x is equal to the sum of even integers from 400 to 600, both inclusive, and y is the number of even integers 400 to 600, both inclusive, what is the value of x+y ?
A)50601 B)50600 C)56002 D)57002 E)50000
sum of even numbers-n^2/4+n/2 applying this the answer is A.
60 sibling pair means 60 students in Junior Class and 60 students in senior class.
P(e)=(60C1/1000C1)*(60C1/800C1)=9/2000 i.e. option C
Pls confirm the OA.
hey Vyom 😃 hwdy ..!!!
Let me explain it in this way ...
let Sibling Set --> all people who have a sibling --> all 120 people
(60C1/1000C1)*(60C1/800C1) --> probability of selecting two people from the Sibling Set (one from each class)and not Siblings
Let (A ,B are Siblings) (C,D are siblings)...
U found the probabilty of picking Aand C from both the classes and not A and B ..!!
Here goes my explanation ...!!
Pick 1 guy from a class of 1000 who belongs to the sibling set
So it s 1 * (60C1/1000C1) ....
Pick one guy from the class of 800 who belongs to he sibling set
Then it is (60C1/800C1) but
these two just belong to the sibling set but not neccesarily siblings Picking one guy from 60 from a the above set will assure you that the guy is the other sibling from the 1000 class =(1/60 )*(60C1/800C1)
So final probability = 1 * (60C1/1000C1) *( 1/60 )*(60C1/800C1)
= 60*60/(60*1000*800)
= 3/40000
P.S: pathetic explanation.....please excuse P.S.S: correct me if i am wrong
let Sibling Set --> all people who have a sibling --> all 120 people
(60C1/1000C1)*(60C1/800C1) --> probability of selecting two people from the Sibling Set (one from each class)and not Siblings
Let (A ,B are Siblings) (C,D are siblings)...
U found the probabilty of picking Aand C from both the classes and not A and B ..!!
Here goes my explanation ...!!
Pick 1 guy from a class of 1000 who belongs to the sibling set
So it s 1 * (60C1/1000C1) ....
Pick one guy from the class of 800 who belongs to he sibling set
Then it is (60C1/800C1) but
these two just belong to the sibling set but not neccesarily siblings Picking one guy from 60 from a the above set will assure you that the guy is the other sibling from the 1000 class =(1/60 )*(60C1/800C1)
So final probability = 1 * (60C1/1000C1) *( 1/60 )*(60C1/800C1)
= 60*60/(60*1000*800)
= 3/40000
P.S: pathetic explanation.....please excuse P.S.S: correct me if i am wrong
Yep. I believe.. I didnt focussed myself on this part of the problem.. Newayz..thanks for highlighting it..
This is exactly what I had. OG 11 has it wrong looks like. As per the book, the OA is: A
The explanation is 60/1000 for juniors and 1/800 for seniors so multiply both for A.
You and I had 60/1000 * 60/800 Which I think is correct.
By the way, you are working out my problems, and I am correcting your SCs in the other thread...totally cool timing. Don't we love PG forum !! :thumbsup:
A certain Junior class has 1000 students, and a certain senor class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A) 3/40000 B) 1/3600 C) 9/2000 D) 1/60 E) 1/15
60 sibling pair means 60 students in Junior Class and 60 students in senior class.
P(e)=(60C1/1000C1)*(60C1/800C1)=9/2000 i.e. option C
Pls confirm the OA.
This is exactly what I had. OG 11 has it wrong looks like. As per the book, the OA is: A
The explanation is 60/1000 for juniors and 1/800 for seniors so multiply both for A.
You and I had 60/1000 * 60/800 Which I think is correct.
By the way, you are working out my problems, and I am correcting your SCs in the other thread...totally cool timing. Don't we love PG forum !! :thumbsup:
Ken left a job paying $75k per year to accept a sales job paying $45k per year plus 15 percent commission.If each of his sales is for $750,what is the least number of sales he must make per year if he is not to lose money because of the change? A-40 B-200 C-266 D-267 E-600
Ken left a job paying $75k per year to accept a sales job paying $45k per year plus 15 percent commission.If each of his sales is for $750,what is the least number of sales he must make per year if he is not to lose money because of the change? A-40 B-200 C-266 D-267 E-600
15% (750 x n) = 75000 - 45000 n=266.666 This is for the bear minimum number of sales. 266 will not be sufficient. Therefore 267.
a juice manufacturer organized taste - testing sessions featuring four brands of orange juice, A, B , C and D. All customers who participated told the organizer which variety they they thought was the best. exactly 61% preerred brand A and exactly half as many prefereed brand B. Only 65 chose brand C. which of the foll could be the number of customers who liked brand D? a> 8 b> 11 c>14 d>20 e>31
1. Heights of women in a city follows a normal distribution with mean 160 cm and standard deviation of 6 cm. In a normal distribution, only 0.0063% of the population is not within 4 standard devistions of the mean. If 5 women are more than 184 cm tall, approximately how many women stay in the city? a>16000 b>40000 c>80000 d> 100,000 e> 160,000
2. How many positive integers less than 100 have exactly 4 odd factors and no even factors?
a>13 b>14 c>15 d>16 e>17
Puys, please explain ur methods the OA are as follws: 1.E 2.C
a juice manufacturer organized taste - testing sessions featuring four brands of orange juice, A, B , C and D. All customers who participated told the organizer which variety they they thought was the best. exactly 61% preerred brand A and exactly half as many prefereed brand B. Only 65 chose brand C. which of the foll could be the number of customers who liked brand D? a> 8 b> 11 c>14 d>20 e>31
Brand A-61% Brand B-30.5% Brand C & D- 8.5%
C is 65 so 8.5% = 65+d
only when d = 20 100% is integer i.e if d=20 than 8.5% is 85 and 100% is (100/8.5)*85 = 1000
Got several methods and several different answers.. 
