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60% of the chocolates in a box are milk chocolate, and 40% are plain chocolate. A third of the milk chocolates and a quarter of the plain chocolates contain nuts. I choose a chocolate at random from the box. Calculate the probability that I have chosen a plain chocolate that does not contain nuts.

Correct me if I am not wrong.

40% of plain chocolates = 0.40 and quarter of this are having nuts means 0.40/4= 0.1 i.e plain chocolates having no nuts =1-0.1 =0.9.

so P(P and No nuts)=P(Plain)*P(No nuts/Plain) i.e 0.4*0.9=0.36. bus this is wrong. I don't know why.

The correct answer is- 0.3

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You simply want the probability that a chocolate randomly selected from the box is a plain chocolate that does not contain nuts. The probability that a chocolate is plain is $0.4$. Since $25\%$ of the plain chocolates contain nuts, $75\%$ do not. Hence, the probability that a chocolate randomly selected from the box is a plain chocolate that does not contain nuts is $P = 0.75 \cdot 0.4 = 0.3$.

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  • $\begingroup$ thanks now I understood. I was doing mistake in calculating quarter of the plain chocolates $\endgroup$ – RajSharma Oct 14 '15 at 13:14
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Given $$\Pr(P)=0.4,\quad \Pr(N|P)=0.25 \implies \Pr(NN|P)=0.75$$

so

$$\Pr(NN \cap P)=\Pr(NN|P)\Pr(P)=(0.75)(0.4)=0.3$$

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