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Numerical Ability
Probability
In a box, there are 3 red balls, 4 green balls and 5 blue balls. Find the probability of drawing 3 balls of different colours one after another without replacement.
Read Solution (Total 20)
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- Total number of balls 12
required probability 3![(3/12)(4/11)(5/10)]=3/11 this is the write answer - 11 years agoHelpfull: Yes(48) No(3)
- 3c1*4c1*5c1/12c3 = 3/11
- 11 years agoHelpfull: Yes(20) No(4)
- i think it is 3/12*4/11*5/10*3=3/22
plz let me knw if i m wrng - 11 years agoHelpfull: Yes(6) No(11)
- i think answer would be 6/22.becoz 3 ball balls can b selected in 3! ways .
- 11 years agoHelpfull: Yes(3) No(1)
- 3*(3/12 * 4/11 * 5/10)=3/22
- 11 years agoHelpfull: Yes(2) No(4)
- (3C1*4C1*5C1)/12C3= 3/11
- 11 years agoHelpfull: Yes(2) No(1)
- 3c1*4c1*5c1/9c3=(3*4*5/9*8*7)*(1*2*3)=5/7
- 11 years agoHelpfull: Yes(1) No(10)
- 3/12*4/11*5/10*3=3/22
- 11 years agoHelpfull: Yes(1) No(6)
- (3c1/12c1*4c1/11c1*5c1/10c1)*3=3/22
- 11 years agoHelpfull: Yes(1) No(1)
- (3/12)(4/11)(5/10) whats the need to multiply with 3!??
- 11 years agoHelpfull: Yes(1) No(0)
- yr galat ans q daal rkha hai. right answer is 41/44 !!
- 11 years agoHelpfull: Yes(1) No(2)
- is it 2/11.?
- 11 years agoHelpfull: Yes(0) No(7)
- 3/12*4/11*5/10*3=3/22
- 11 years agoHelpfull: Yes(0) No(3)
- 1st ball is red then probability is 1/3
2nd ball is green then probability is 1/4
3rd ball is red then probability is 1/5
total probability=((1/3)*(1/4)*(1/5))/(12c3) - 11 years agoHelpfull: Yes(0) No(1)
- 6*3/12*5/11*4/10=3/11
- 11 years agoHelpfull: Yes(0) No(1)
- 3/11...............................
- 11 years agoHelpfull: Yes(0) No(0)
- ans. is 15/22.Since balls are drawn one after another so
3C1/12C1 + 4C1/11C1+ 5C1/10 ==15/22 - 11 years agoHelpfull: Yes(0) No(0)
- 3c1*4c1*5c1/12c3 = 3/11
is the right ans - 11 years agoHelpfull: Yes(0) No(0)
- answer is 1/22
- 11 years agoHelpfull: Yes(0) No(0)
- i think ans will be 7/22
1-((3!+4!+5!)/12c3) - 11 years agoHelpfull: Yes(0) No(0)
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