Official Quant thread for CAT 2013

MBA CAT 2022
2681
@ani4588 said:
A standard pack of playing cards has 52 cards, divided into 4 suits of 13 cards each having 13 different values from 1 to 13. Three cards are drawn from such a pack after sufficient shuffling. Find the probability that all three are of different suits and different values.
(4c3 x 13 x 12 x 11 ) / 52c3

?
2682
@ani4588 said:
A standard pack of playing cards has 52 cards, divided into 4 suits of 13 cards each having 13 different values from 1 to 13. Three cards are drawn from such a pack after sufficient shuffling. Find the probability that all three are of different suits and different values.
132/425 or .3106?????
2683
@ani4588 said:
A standard pack of playing cards has 52 cards, divided into 4 suits of 13 cards each having 13 different values from 1 to 13. Three cards are drawn from such a pack after sufficient shuffling. Find the probability that all three are of different suits and different values.
sample space = 52C3 = 22100
favourable cases = 13C3 * 4P3 = 6864

probability = 6864/22100 = 1716/5525
2684
@ani4588 said:
A standard pack of playing cards has 52 cards, divided into 4 suits of 13 cards each having 13 different values from 1 to 13. Three cards are drawn from such a pack after sufficient shuffling. Find the probability that all three are of different suits and different values.
132/425.?

Correct me if I'm wrong
2685
@sbharadwaj @gs4890 @grkkrg @19rsb OA is 132/425
2686
@krum said:
b1+b2+b3=299total=298c2b1=150 => 148b1=149 => 147...b1=297 => 1p=[148/2*149]/298c2=11026/44253=0.25
Please anyone explain where I'm wrong. :/

Question was

A box B1 has 299 identical cards. These cards are divided into three heaps and placed in boxes B1, B2 and B3 such that each box has at least one card. What is the probability that B1 has more cards than B2 and B3 put together?
My approach

Total outcomes
Total 299 cards
Give one to all three(atleast one card should be there) remaining will be 296
Divide 296 into group of three 298c2 = total outcomes
---
Favourable outcomes
Total 299 cards
b1>b2+b3
b1 should have 150 cards with him
remaining 149 are to be divided amoung b2 and b3
Give one to both(atleast one card should be there)
Remaining 147 has to be divided amoung 2
148c1= 148 = favourable (which is wrong)
2687
@ani4588 said:
A standard pack of playing cards has 52 cards, divided into 4 suits of 13 cards each having 13 different values from 1 to 13. Three cards are drawn from such a pack after sufficient shuffling. Find the probability that all three are of different suits and different values.
Select 3 out of the 4 colors in 4C3 ways
so 4C3*13*12*11/52C3 = 132/425
2688
@19rsb said:
is it 610????
Hmm I guess.!!
2689
@Torque024 said:
Please anyone explain where I'm wrong. :/Question wasA box B1 has 299 identical cards. These cards are divided into three heaps and placed in boxes B1, B2 and B3 such that each box has at least one card. What is the probability that B1 has more cards than B2 and B3 put together?My approachTotal outcomesTotal 299 cardsGive one to all three(atleast one card should be there) remaining will be 296Divide 296 into group of three 298c2 = total outcomes---Favourable outcomesTotal 299 cardsb1>b2+b3b1 should have 150 cards with himremaining 149 are to be divided amoung b2 and b3Give one to both(atleast one card should be there)Remaining 147 has to be divided amoung 2148c1= 148 = favourable (which is wrong)
b1 can have 151,152,153..297 cards

each of the case u have to find no.of ways of arranging cards in b2 and b3..


if u do that u get a series whose sum is 148*149/2
2690
@wovfactorAPS said:
b1 can have 151,152,153..297 cardseach of the case u have to find no.of ways of arranging cards in b2 and b3..if u do that u get a series whose sum is 148*149/2
Ya, aa gaya samaj. A ko 150 dene ke baad Remaining 147 has to be divided among 3 = 149c2
Thanku :)
2691

@Torque024 B1 need not have 150... it can have 101 also

101+100+98 wil also do... so i gues u have missed such cases..

2692

oh yeah.. above cases where b1 can be more than 151 etc.. tat was what i wanted to tell but told something else

2693
2^29 is a 9 digit number containing all distinct digits. Which number is missing ?
2694
Recently smoking at public places is declared as an offence. Delhi Police has started imposing
a penalty against smoking in public and has eight raid teams in place. In a surprise check, the
raid team caught 40 people smoking in the Connaught Place area of Delhi. The standard
deviation and sum of squares of the amount found in their pockets were Rs. 10 and Rs. 40000,
respectively. If the total fine imposed on these offenders is equal to the total amount found in
their pockets and the fine imposed is uniform, what is the amount that each offender will have
to pay as fine?
(1) Rs. 90(2) Rs. 60(3) Rs. 30(4) Rs. 15
OA is rs30....share ur approach also.....plz......
2695
@rkshtsurana said:
2^29 is a 9 digit number containing all distinct digits. Which number is missing ?
is it 4?????
2696
@19rsb yup 😛 shre d approach
2697
@rkshtsurana said:
2^29 is a 9 digit number containing all distinct digits. Which number is missing ?
4.8 = 32 = 5 remainder with 9
so 4 is missing here.
2698
@sujamait said:
4.8 = 32 = 5 remainder with 9so 5 is missing here.
4 is the remainder wid 9 ..
2699
@rkshtsurana said:
4 is the remainder wid 9 ..
5 hai bha iremainder.
so 4 mising ho jayega..tabhi 5+4 = 9 banega 1
2700
@sujamait said:
5 hai bha iremainder.
oh sry 😞 ..but ans is 4..