thanks krusta for your input. i really appreciate your efforts (so that i know im not wasting my time posting questions)
these are the answers. made up the q myself.
84) There are 6 different colored balls along with 6 corresponding colored boxes.
a) What is the probability that exactly 1 ball will go into its corresponding box?
6 * 1/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1/1 = 1/5
Explanation
6 balls --> call them Ball 1, Ball 2 …etc
6 boxes
Assume that Ball 1 goes into its CORRECT corresponding box
Ball 1 has 1 possible CORRECT box out of 6 boxes.--> 1/6
Now every subsequent ball has to be WRONG.
Ball 2 has 1 CORRECT box of 5 boxes.
Ball 2 has 4 WRONG boxes out of 5 boxes --> 4/5
Ball 3 goes in wrong box--> 3/4
Ball 4 goes in wrong box--> 2/3
Ball 5 goes in wrong box-->1/2
Ball 6 goes in wrong box-->1/1
Joint probability = 1/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1/1 = 1/5
This joint probability is where only Ball 1 is correctly placed.
The stem did not say which ball is correctly placed.
Therefore, we must multiply the joint probability by 6 to get all 6 possible scenarios.
Therefore, 6 * 1/6 * 4/5 * 3/4 * 2/3 * 1/2 * 1/1
b) What is the probability that exactly 2 balls will go into its corresponding box?
6 * 1/6 * 1/5 * 3/4 * 2/3 * 1/2 * 1/1 = 1/20
c) What is the probability that exactly 5 balls will go into its corresponding box?
6 * 1/6 * 1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/120
d) What is the probability that all balls will go into its corresponding box?
Ball 1 goes into box 1, ball 2 goes into box 2…etc.
This scenario let’s all the balls go in their respective boxes.
We don’t multiply by 6 because each ball goes exactly where its supposed to. Whether ball 1 goes in the box 1 first or ball 2 goes into box 2 first, the final outcome is the same.
1/6 * 1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/720
1/6! = 1/720
e) What is the probability that no balls will go into its corresponding box?
1 – all = none
1 – 1/6! = 719/720