Latest TCS Aptitude Question SOLUTION: A drawer has 4 red hat & 4 blue hat .find d prblty f getting xatly 3 blue hats when taking out 4 hats randomly out f d drawer and immediatly returning every hat to the drawer befor

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(#M40039566) TCS Numerical Ability question Keep an EYE

A drawer has 4 red hat & 4 blue hat .find d prblty f getting xatly 3 blue hats when taking out 4 hats randomly out f d drawer and immediatly returning every hat to the drawer before taking out the next???

Asked In TCS ranjan kumar (9 years ago)

Unsolved Read Solution (22)

Is this Puzzle helpful?   (16)   (1) Submit Your Solution Probability

Read Solution (22) :

suraj jha  (9 years ago)

(4C3*4C1)/8C4

Is this solution Helpfull? Yes (28) | No (9)  

akash tiwari  (9 years ago)

P(BBBR) = (1/2)^4
P(BBRB) = (1/2)^4
P(BRBB) = (1/2)^4
P(RBBB) = (1/2)^4
total = 4 * (1/2)^4
ans :1/4

Is this solution Helpfull? Yes (21) | No (2)  

BharathVattikuti  (9 years ago)

probability for taking first blue hat is 4/8
second one also 4/8
for third one also 4/8
now the fourth hat must be red because they mentioned they require exactly 3 blue hats.
probability for red is also 4/8
so in total = (4/8)^4

Is this solution Helpfull? Yes (7) | No (3)  

manaswini devabhakthuni  (9 years ago)

4c3*4c1/(8c4)=8/35

Is this solution Helpfull? Yes (4) | No (1)  

bhaskar  (9 years ago)

first of all we have to take one blue hat out of 8 hats which can be done in 4c1/8c1 ways.
and then we return it to drawer and take another blue hat in 4c1/8c1 ways
and then we return it to drawer and take another blue hat in 4c1/8c1 ways
next 4th hat is taken in 8c1/8c1 ways
so total probability is (4c1*4c1*4c1*8c1)/(8c1*8c1*8c1*8c1)

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

Ajay Singh Yadav  (9 years ago)

p(r)=1/4 p(b)=1/4
now our objective=bbbr*4!/3!
as blue hats are returned back so p(b) remains 1/4
our ans=1/4*1/4*1/4*1/4*4!/3!=1/64

Is this solution Helpfull? Yes (2) | No (1)  

jawed  (9 years ago)

1/5

(1/2*1/2*1/2)/(1/2*1/2*1/2+1/2)

Is this solution Helpfull? Yes (1) | No   

geetanjali  (9 years ago)

(4c3*4c1)/8c4

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

Manjunath  (9 years ago)

P(BBBR) = (1/2)^4
P(BBRB) = (1/2)^4
P(BRBB) = (1/2)^4
P(RBBB) = (1/2)^4
total = p(BBBR)+P(BBRB)+P(BRBB)+P(RBBB)=1/4
ans :1/4

Is this solution Helpfull? Yes (1) | No   

shiva  (9 years ago)

sabse pehla wala galat hai soln

Is this solution Helpfull? Yes (1) | No (2)  

ANKIT SHARMA  (9 years ago)

((1/4)^4)*4/8c1 multiply 4 becoz 4cases exits

Is this solution Helpfull? Yes | No   

sahithi  (9 years ago)

to take a blue hat - 4c1/8c1.Give it immediately.
another blue hat - 3c1/7c1,give it.
another - 2c1/6c1.
Exactly three blue hats done and given.
Now 4th hat to be red hat - 4c1/5c1 (5c1 bcoz,3blue hats given and 4 red n 1blue hat are the remaining ones)
So, (4c1/8c1) * (3c1/7c1) * (2c1/6c1) * (4c1/5c1)

Is this solution Helpfull? Yes | No (3)  

rahul gera  (9 years ago)

4c3*4c1/8c4

Is this solution Helpfull? Yes | No   

Arnab Mitra  (9 years ago)

(4c1/8c1)*4=1/16
now 3 exactly blue hats can be obtained in following ways(bbbr),(bbrb),(brbb),(rbbb)so 4*(1/16)=1/4.

Is this solution Helpfull? Yes | No   

vivek kumar  (9 years ago)

RRRB HAT CAN BE ARRANGED IN 4 WAYS BECAUSE PERMUTATION OF RRRB IS 4!/3!=4
PROBABILITY=RRRB+RRBR+RBRR+BRRR
= 4*(4*4*4*4)/(8*8*8*8)=1/4

Is this solution Helpfull? Yes | No   

mounika  (9 years ago)

plz any one say what is crct ans for this.

Is this solution Helpfull? Yes | No   

Raghav gupta  (9 years ago)

(4c1*4c1*4c1*8c1)/(8c1*8c1*8c1*8c1)

Is this solution Helpfull? Yes | No   

Amarpreet Singh  (9 years ago)

4C1*4C1/8C1

Is this solution Helpfull? Yes | No   

saptarsi ghosh  (9 years ago)

4c1*4c1*4c1*4c1/8c4=128/35

Is this solution Helpfull? Yes | No   

milan jena  (8 years ago)

We can choose in (3,1)and(1,3) ways.
so for 1st condition it is (4/8)*(4/8)*(4/8)*(4/8)*(4/8)=1/16
similarly in second condition it is also 1/16
so total probability is 1/16 + 1/16=2/8

Is this solution Helpfull? Yes | No   

NARENDRA KUMAR  (8 years ago)

Two types solution exists here:
1)
if all balls are identical then
8c3/10!

2)
if all balls are not identical :
7!*8c3*3!/10!

Is this solution Helpfull? Yes | No (1)  

vivek kumar gupta  (8 years ago)

When you take out hats one by one after replacement, there are equal chances of getting red or blue hat.

So possible outcomes are
RRRB, RRBR,RBRR,BRRR, BBBR,BBRB,BRBB,RBBB.. with exactly three red or blue hats.
RRBB,BBRR,RBRB,BRBR,BRRB,RBBR, BBBB,RRRR.. with other combinations.

so out of 16 possible combinations , eight are desired combinations.
so probability of getting exactly 3 red hats or exactly 3 blue hats when taking out 4 hats at random and return every hat to drawer before taking out another one = 8/16 =1/2

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