Latest TCS Aptitude Question SOLUTION: There are 6 red, 8 blue, 7 green in a bag. If 5 are drawn with replacement what is probability at least 3 are red

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(#M40041085) TCS Numerical Ability question probability Keep an EYE

There are 6 red, 8 blue, 7 green in a bag. If 5 are drawn with replacement what is probability at least 3 are red

Asked In TCS Sonal jain (9 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

srikanth  (9 years ago)

(6c3*15c2)/21c5 + (6c4 * 15c1)/21c5 + (6c5 * 15c0)/21c5

Is this solution Helpfull? Yes (24) | No (2)  

soubhagya  (9 years ago)

total balls 6+8+7=21
5 are drawn at a time
at least 3 red balls must be there out of 5 balls
so the there must be 3 or 4 or 5 red balls
for 3 red balls-( 6/21 *6/21*6/21*15/21*15/21) here 15/21 bcoz we have to choose 2 balls out of 15 rest balls
like this if we proceed for 4 balls and 5 balls and at last add them then we got the result.

Is this solution Helpfull? Yes (9) | No   

Rahil  (9 years ago)

total balls 6+8+7=21
5 are drawn at a time
at least 3 red balls must be there out of 5 balls
so the there must be 3 or 4 or 5 red balls
for 3 red balls-( 6/21 *6/21*6/21*15/21*15/21) here 15/21 bcoz we have to choose 2 balls out of 15 rest balls
like this if we proceed for 4 balls and 5 balls and at last add them then we got the result.
i.e. nothing but.. (6c3*15c2)/21c5 + (6c4 * 15c1)/21c5 + (6c5 * 15c0)/21c5
=2331/20349
which is 37/323

Is this solution Helpfull? Yes (7) | No   

ARCHIE GUPTA  (9 years ago)

2331/20349
which is 37/323

Is this solution Helpfull? Yes (4) | No   

Nilay Saxena  (9 years ago)

for 3 red balls-(5c3)*( 6/21 *6/21*6/21*15/21*15/21)
for 4 red balls-(5c4)*( 6/21 *6/21*6/21*6/21*15/21)
for 5 red balls-(5c5)*( 6/21 *6/21*6/21*6/21*6/21)

add them and we got the result

Is this solution Helpfull? Yes (3) | No   

radha  (9 years ago)

atlest 3red
3red 4red 5red
(6/21)^3*(15/21)^2+(6/21)^4*(15/21)+(6/21)^5

Is this solution Helpfull? Yes (1) | No   

tirupathi reddy  (9 years ago)

259/2261
P(at least three red) =
((6c3*15c2)/21c5 + (6c4 * 15c1)/21c5 + (6c5 * 15c0)/21c5)

Is this solution Helpfull? Yes | No (2)  

harshdeep kumar  (9 years ago)

5c3*(2/7)^3 *(5/7)^2 + 5c4*(2/7)^4 *(5/7) + 5c5*(2/7)^5

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