A probability problem by maryam kouram

since one box shall remain empty, and we have 5 balls, 3 boxes will contain 1 ball each and 1 box will contain two balls. first, we need to calculate the number of possibilities in general, in this situation we have that number equal to5^{5}= 3125.

the number of ways to choose a box that remains empty is 5. the number of ways to choose a box that contains two balls is 4 ( since we have already chosen a box that will remain empty we only have 4 left to deal with ). the number of ways to choose two balls to put them in that box is: 5C2 = 10 and the number of ways to put the 3 balls that we have left in the 3 remaining boxes so that each one contains 1 ball is 3!=6 so CARD(A)=5 $\times 4 \times 10 \times 6$ = 1200 so the number of possibilities is P(A) = $\frac{1200}{3125}$ = 0.384

Since any of the $5$ balls can go into any one of the $5$ boxes, the total number of possible elementary outcomes (sequence of labels of boxes where the balls go) is $5^5$ . Each one of these outcomes is equally likely.

Now, in these outcomes, one box will remain empty if and only if two balls go to a single box and the other three balls go to three distinct boxes, leaving one empty. The total number of such possibilities is $(5 \times \binom{5}{2})\times ^4P_3$ .

Hence the required probability is $\frac{(5 \times \binom{5}{2})\times ^4P_3}{5^5}= 0.384.$