# How to solve this probability question that involves conditional probability?

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.

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 $P = 0.75 \cdot 0.4 = 0.3$.
Given $$\Pr(P)=0.4,\quad \Pr(N|P)=0.25 \implies \Pr(NN|P)=0.75$$
$$\Pr(NN \cap P)=\Pr(NN|P)\Pr(P)=(0.75)(0.4)=0.3$$