Latest TCS Aptitude Question SOLUTION: A seven digit number is formed by using the digits 3, 3, 4, 5, 5, 8 & 9. Probability that the number formed is great than 57,26,000 is? OPTION 1) 3/8 2) 1/5 3) 1/3 4) 2/9 5) 12/35

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A seven digit number is formed by using the digits 3, 3, 4, 5, 5, 8 & 9. Probability that the number formed is great than 57,26,000 is?
OPTION 1) 3/8
2) 1/5
3) 1/3
4) 2/9
5) 12/35
6) 16/35
7) 8/35
8) 8/21
9) 2/5

Asked In TCS Pradeep Saw (9 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

ramakantkumar  (9 years ago)

only four condition...
58........=====> 3 is repeated and first two no is fixed..then..5!/2....

59.........=====>3 is repeated and first two no is fixed..then..5!/2....

8.......=> in this case both 3 and 5 is repeated..so..= 6!/(2*2)

9......===>.in this case both 3 and 5 is repeated..so..= 6!/(2*2)
lower part is 7!/(2*2),,because of repeatation....
so result==>uper part sum..==5!/2 + 5!/2 + 6!/(2*2) + 6!/(2*2)
===> 5!*4
then result===5!*4/(7!/2*2)==5!*4*4/7!==4*4/(7*6)==8/21 Ans..

Is this solution Helpfull? Yes (23) | No (2)  

akshay  (9 years ago)

total sample space is 7!/(2!*2!)=1260
total number of numbers formed by 3,3,4,5,5,8,9 which are greater than 5726000 are
when first digit is 5 second digit should be 8 or 9
so 58*****=5!/2!=60
and 59*****=5!/2!=60
or first digit should be 8 or 9
so 8******=6!/(2!*2!)=180
and 9******=6!/(2!*2!)=180
so required probability = (60+60+180+180)/1260=480/1260=8/21

Is this solution Helpfull? Yes (7) | No (1)  

navin kr prasad  (9 years ago)

ans is 1/3
total no.of outcome=7!=7*6*5*4*3*2*1
case 1 :5 8 _ _ _ _ _
or
case 2 :5 9 _ _ _ _ _
or
case 3:8 _ _ _ _ _ _
or
case 4:9 _ _ _ _ _ _
these 4 cases are favourable no. of outcome
there fore required probability=5!/7!+5!/7!+6!/7!+6!/7!=1/3 ans

Is this solution Helpfull? Yes (5) | No (11)  

goutham  (9 years ago)

8/21
soln:
- - - - - - - the no of ways we can the dashes with these numbers (3, 3, 4, 5, 5, 8 & 9.) is 7!/(2!*2!)=1260
The number of numericals greater than 57,26,000 is 58-----(5!/2!) or 59-----(5!/2!) or 8------(6!/(2!*2!)) or 9------(6!/(2!*2!))
sum of all these is 480
probability = 480/1260= 8/21

Is this solution Helpfull? Yes (3) | No   

AARJU GANGWAR  (9 years ago)

Absolutely Correct Solution......
Total no. of numbers formed by using the given digits 3,3,4,5,5,8,9 = (7*6*5*4*3*2*1)/(2*1)*(2*1) = 1260
(Denominator part is due to the repetition of 3 and 5)
Total no. of numbers formed by using the given digits 3,3,4,5,5,8,9 which are greater than 57,26,000 are of two types
Total no. of numbers started with 5 = (2*2*5*4*3*2*1)/ (2*1)*(2*1) = 120
Total no. of numbers started with 8&9 = (2*6*5*4*3*2*1)/ (2*1)*(2*1) = 360
Hence required probability = (120 + 360)/ 1260 = 480/1260 = 8/21

Is this solution Helpfull? Yes (3) | No   

Mudrika Dhanda  (9 years ago)

please provide proper solution

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pravartika  (9 years ago)

according to me . no of favourable outcomes would be 5!+5! cz the no greter than 5726000 would hav either 58 in thebegining or 59 in the begining

total no of outcomes would be 7!

probability =2*5!/ 7!
=1/21

Is this solution Helpfull? Yes | No (4)  

Parvez khan  (9 years ago)

total no. of 7 digit no.s = 7!/(2!*2!)=1260
case 1 : 5,8 _ _ _ _ _=240/4=60
case 2 : 5,9_ _ _ _ _ =240/4=60
case 3 : 8,9,_ _ _ _ _ =120/4=30
case 4: 9,8,_ _ _ _ _ =120/4=30
p= (60+60+30+30)/1260=1/7

Is this solution Helpfull? Yes | No (1)  

K.L.Manasa  (9 years ago)

There can be another case.. 57-------.
THEN (7*6*5*4*3/2*2)/(7!/2*2)+8/21

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Antriksha  (9 years ago)

- - - - - - - the no of ways we can the dashes with these numbers (3, 3, 4, 5, 5, 8 & 9.) is 7!/(2!*2!)=1260
The number of numericals greater than 57,26,000 is 58-----(5!/2!) or 59-----(5!/2!) or 8------(6!/(2!*2!)) or 9------(6!/(2!*2!))
sum of all these is 480
probability = 480/1260= 8/21

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