Latest TCS Aptitude Question SOLUTION: A basket contains 6 red balls, 5 blue balls, 4 green balls and 3 white balls. Five balls are to be drawn together at random, then what is the probability that there is at least a b

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(#M40039492) TCS Numerical Ability question probability Keep an EYE

A basket contains 6 red balls, 5 blue balls, 4 green balls and 3 white balls. Five balls are to be drawn together at random, then what is the probability that there is at least a ball of each colour?

a. 10/51
b. 5/51
c. 10/17
d. 5/17

Asked In TCS shalini agrawal (9 years ago)

Unsolved Read Solution (18)

Is this Puzzle helpful?   (16)   (4) Submit Your Solution Probability

Read Solution (18) :

abhay singh  (9 years ago)

required probability={(6*5*4*3c2)+(6*5*4c2*3)+(6*5c2*4*3)+(6c2+5*4*3)}/18c5
=(360+540+720+900)/18*17*14*2
=5/17

Is this solution Helpfull? Yes (34) | No (2)  

virendra tiwari  (9 years ago)

6c2*5c1*4c1*3c1 + 6c1*5c2*4c1*3c1 + 6c1*5c1*4c2*3c1 + 6c1*5c1*4c1*3c2 =2520
total 18c5
ans 25208568= 5/17

Is this solution Helpfull? Yes (23) | No   

sujit kumar jha  (9 years ago)

as we have to draw 5 balls in which at leat 1 ball of each colour should be present..so it can be done in four ways...by selecing 2 balls from one colour and rest 3 from remaining 3 colours...so required probability=
{(6c2*5c1*4c1*3c1)+(6c1*5c2*4c1*3c1)+(6c1*5c1*4c2*3c1)+(6c1*5c1*4c1*3c2)}/18c5
=5/17

Is this solution Helpfull? Yes (8) | No   

Aritra Das  (9 years ago)

there are total 5 balls selected and we have to select at least one from each color, that leaves that the remaining ball can be of either of 4 colors
there will be 4 cases
case 1 : last ball is red
6c2*5c1*4c1*3c1
case 2 : blue
6c1*5c2*4c1*3c1
and this will go like this for rest 2 cases

now we can select 5 balls from 18 in 18c5 ways
so the required probability will be
(6c2*5c1*4c1*3c1 + 6c1*5c2*4c1*3c1 + 6c1*5c1*4c2*3c1 + 6c1*5c1*4c1*3c2)/18c5
= 5/17

Is this solution Helpfull? Yes (5) | No   

Narasapu Hari prasad  (9 years ago)

total reqiured possible ways are (6*5*4*3c2)+(6*5*4c2*3)+(6*5c2*4*3)+(6c2*5*4*3)===>>>360+540+720+900=2520
total ways of selecting 5 balls=18c5=8568
so probability=2520/8568=5/17

Is this solution Helpfull? Yes (3) | No   

Pranoy Bej  (9 years ago)

ANSWER ==> 5/17
P(E)= {(6C1*5C1*4C1*3C2)+(6C1*5C1*4C2*3C1)+(6C1*5C2*4C1*3C1)+(6C2*5C1*4C1*3C1)}/18C5 = 2520/8568 = 5/17

Is this solution Helpfull? Yes (3) | No   

Govind raj  (9 years ago)

p(e)=(3c2*6c1*5c1*4c1)+(3c1*6c2*5c1*4c1)+(3c1*6c1*5c2*4c1)+(3c1*6c1*5c1*4c2)/(18c5)=10/17 ans....

Is this solution Helpfull? Yes (3) | No   

Aastha  (9 years ago)

probability=[(6C2*5*4*3)+(6*5C2*4*3)+(6*5*4C2*3)+(6*5*4*3C2)]/18C5
on solving =5/17

Is this solution Helpfull? Yes (2) | No   

DEEPANKUMAR  (9 years ago)

probability=((6c2*5c1*4c1*3c1)+(6c1*5c2*4c1*3c1)+(6c1*5c1*4c2*3c1)+(6c1*5c1*4c1*3c2))/18c5
=5/17

Is this solution Helpfull? Yes (1) | No   

pooja chavan  (9 years ago)

(6c2*5c1*4c1*3c1+6c1*5c2*4c1*3c1+6c1*5c2*4c1*3c1+6c2*5c1*4c1*3c1)/18c5=5/17

Is this solution Helpfull? Yes (1) | No   

ganesh pache  (9 years ago)

Ans=c
n(s)=18C5=8568
n(a)=6c1*5C1*4c1*3c1*14c1=5040
p9a0=5040/8658

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

Sahil Chawla  (9 years ago)

D
taking 4 cases with 2 balls frm red,blue,green,white resp. in each case nd rest others balls of other clrs

Is this solution Helpfull? Yes | No   

vishnudeekshit  (9 years ago)

(6c1*5c1*4c1*3c1)(14c1)/18c5=10/17

Is this solution Helpfull? Yes | No (5)  

ARPITA AICH  (9 years ago)

As only 2 types of socks exist. Let us assume for simplicity 5pair are red and 5pair are blue.
total socks 20 and we have to select 3. then n(s)=20C3
we can select one pair of red sock and 1 blue sock or 1 pair of blue sock and 1 red sock
two combination possible
hence (10C2*10C1+10C2*10C1)/20C3
ANS 900/1140=15/19

Is this solution Helpfull? Yes | No (3)  

akash tiwari  (9 years ago)

6*5*4*3*14c1/18c5

Is this solution Helpfull? Yes | No   

rahul gera  (9 years ago)

6C2*5*4*3+6*5C2*4*3+6*5*4C2*3+6*5*4*3C2/18C5
=5/17

Is this solution Helpfull? Yes | No   

Arun kumar  (9 years ago)

{(6*5*4*3c2)+(6*5*4c2*3)+(6*5c2*4*3)+(6c2+5*4*3)}/18c5
=(360+540+720+900)/18*17*14*2
=5/17

Is this solution Helpfull? Yes | No   

SK IMRAN   (9 years ago)

can any one explain why the following process is not correct??

(6c1*5c1*4c1*3c1)(14c1)/18c5=10/17

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