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Consider an urn containing 5 red,5 black and 10 white balls.If balls are drawn without replacement from the urn,calculate the probability that in the first 7 draws, at least one ball of each colour is drawn.

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Let Rc: denote the event that there are no red balls in the first 7 draws.

Let Bc: denote the event that there are no black balls in the first 7 draws.

Let Wc: denote the event that there are no white balls in the first 7 draws.

P(at least one ball of each color is drawn in the first 7 draws)= P(Rc’Bc’Wc’)=1-P(Rc U Bc U Wc)

Now to calculate P(Rc U Bc U Wc)= P(Rc)+P(Bc)+P(Wc)-P(RcBc)-P(BcWc)-P(WcRc)+P(RcBcWc)

Note that: Balls are drawn without replacement from the urn.

P(Rc)=P(no red balls in the first 7 draws)= 15C7/20C7

P(Bc)=P(no black balls in the first 7 draws)=15C7/20C7

P(Wc)=P(no white balls in the first 7 draws)=10C7/20C7

P(RcBc)=P(no red and black balls in the first 7 draws)=10C7/20C7

P(BcWc)=P(no black and white balls in the first 7 draws)=0

P(WcRc)=P(no white and red balls in the first 7 draws)=0

P(RcBcWc)=P(no red and white and black balls in the first 7 draws)=0

Hence P(Rc U Bc U Wc)= 2*(15C7)/(20C7)

Therefore P(Rc’Bc’Wc’)= 1- 2*(15C7)/(20C7)= 0.834 (approx).

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