Latest TCS Aptitude Question SOLUTION: From a collection of 12 bulbs of which 6 are defective 3 bulbs are chosen at random for three sockets in a room.Find the probability that the room is lighted if one bulb is suffic

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(#M40152813) TCS Numerical Ability question Probability Keep an EYE

From a collection of 12 bulbs of which 6 are defective 3 bulbs are chosen at random for three sockets in a room.Find the probability that the room is lighted if one bulb is sufficient to light the room.

Asked In TCS Anura Jain (9 years ago)

Unsolved Read Solution (11)

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Read Solution (11) :

srikanrh  (9 years ago)

1-(6C3/12C3) = 10/11

Is this solution Helpfull? Yes (11) | No   

tina  (9 years ago)

Prob= [6c3*6c0+6c2*6c1+6c2*6c1]/[12c3]=10/11

Is this solution Helpfull? Yes (9) | No   

RAGNI RUPAM  (9 years ago)

total ways of choosing three bulbs=12C3
if 0 defective bulb is choosen= 6C3
if 1 ...........................................= 6C1*6C2
if 2 .......................................... = 6C2*6C1
total ways in which room room is lighted = 6C3+(6C1*6C2)+(6C2*6C1)
reqd prob= [6C3+(6C1*6C2)+(6C2*6C1)]/12C3 = 10/11
=> reqd prob= 10/11

Is this solution Helpfull? Yes (5) | No   

vidya bharati  (9 years ago)

To lighten the room at least one non defective ball is required:
ways of selecting at least one non defective ball,lets say n= (6C1 * 6C2)+(6C2*6C1)+6C3
total ways of selecting 3 ball ,lets say s=12C3
probability=n/s= 10/11

Is this solution Helpfull? Yes (1) | No   

Mohd Shoeb  (9 years ago)

prob= 10/11

Is this solution Helpfull? Yes (1) | No   

radha shaw  (9 years ago)

2*(6C1*6C2/12C3)+(6C3/12C3)
=7/11

Is this solution Helpfull? Yes | No (3)  

DEEPAK  (9 years ago)

room can be light in 3 ways
if(i good bulb select) or (2 good bulb selected) or (all good bulb selected)
(1/2) + (1/2*5/11) + (1/2*5/11*2/5)=9/11 ans

Is this solution Helpfull? Yes | No (4)  

Amandeep Gupta  (9 years ago)

(6C1+6C2+6C3)/12C3 ... simply take the probability of each case of taking 1 bulb, 2 and 3 out separately and add them ..

Is this solution Helpfull? Yes | No (1)  

Aparna Takkellapati  (9 years ago)

Prob =10/11

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Sarbapriya Panja  (9 years ago)

total no of combination=12c3=220
case1, 1bulb is on=11c2=55
case2,2 bulbs on=10c1=10
case3, 3bulbs on=6c3=20
total favourable case=85
p=85/220=17/44

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siva kumar  (8 years ago)

total number of chances,n(S)=12;
chances of lighting room n(A)=6;
therefore probability of lighting the room= 6/12=1/2

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