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A bo x co ntains 2 white balls 3 blac k balls and 4 red balls. I n how many ways can 3
balls be drawn fro m the box , if at least one blac k ball is to be included in the draw ?
Read Solution (Total 10)
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- (3c1*6c2)+(3c2*6c1)+1=64
- 12 years agoHelpfull: Yes(17) No(3)
- (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
= (45 + 18 + 1)= 64. - 12 years agoHelpfull: Yes(8) No(0)
- 3c1*6c2+3c2*6c1+3c3
- 12 years agoHelpfull: Yes(2) No(0)
- see this is the answer calculate it (3c1*6c2)/(9c3)
- 12 years agoHelpfull: Yes(1) No(4)
- n(S)=no of ways of drawing 3 balls out 9c3=84
E=event of drawing 1 black
n(E)=3c1=3
p(E)=n(E)/n(S)=3/84=1/28 (ans) - 12 years agoHelpfull: Yes(1) No(7)
- Snehashish is right...
- 12 years agoHelpfull: Yes(1) No(0)
- oh!! i forget that ..thnxx SHYAMLESH
- 12 years agoHelpfull: Yes(1) No(0)
- see i am correct ans is 15/28 u have to consider the ways in which other balls are being choose so in that case its one from black and the rest two from other set of balls
- 12 years agoHelpfull: Yes(0) No(2)
- 30
since atleast 1 black ball to be drawn1 roblack n other from g other balls
similarly 2 from black and 3 from black
1*6c2+2*6c1+3*6c0=30
- 12 years agoHelpfull: Yes(0) No(1)
- pls give us the choices to this question as it is confusing 2 ways which way to go sample space has to be divided or not and if the replacemnt is dere or randomly drawn
- 12 years agoHelpfull: Yes(0) No(0)
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