Explanation
$$\\\>P(\frac{B}{A})=\>\frac{P(B\>\cap\>A)}{P(A)}\\\>0.4=\frac{P(B\cap\>A)}{0.8}\\$$
$$\therefore\>P(B\cap\>A)=0.32\\$$, $$\implies \>P(A\cap\>B)=0.32\\$$
$$then\>\>P(\frac{A}{B})=\>\frac{P(A\>\cap\>B)}{P(B)}=(\frac{0.32}{0.5})=0.64\>\\$$
$$and\>P(A\cup\>B)=P(A)+P(B)-P(A\cap\>B)\\=0.8+0.5-0.32\\\>=0.98$$
A & B are two sets
AᴜB = (A-B) + (A∩B) + (B-A)
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