Q.(Activity) A bag contains 3 red and 5 white balls. Two balls are drawn at random one after the other without replacement. Find the probability that both the balls are white. (Complete the following activity.)

Bag: 3 red + 5 white = 8 balls. Let $A$ = first ball white, $B$ = second ball white. $P(A)=5/8$. After removing one white ball (not replaced), 7 balls remain: 3 red, 4 white, so $P(B/A)=4/7$. By the multiplication theorem, $P(\text{both white})=P(A)\cdot P(B/A)=\frac{5}{8}\times\frac{4}{7}=\frac{20}{56}=\frac{5}{14} …

Multiplication theorem, filling in the activity's blanks: first probability, reduced ball count, …

Filling the 'remaining balls' blank as 8 instead of 7, forgetting the first (unreplaced …