If we have 3 types of things that were given for 4 people what the probability that all people will select the same thing, if they can select what they want
Or have to get more information to solve this problem? More information in the problem given 32 things that have 3 types (A,B,C) A - 10 things B - 8 things C - 14 things
Thanks for answer
The problem is not solvable unless each person selects items at random. Also, certainly, one does need to know how many items there are of each type. Still the statement is a bit vague. I suppose each person will pick exactly one item.
It seems you want $P(\text{all 4 select } A) + P(\text{all 4 select }B) + P(\text {all 4 select }C).$ Does that seem correct? If so, what rule of probability is involved?
Then if items are selected without replacement $$P(\text{all 4 select} A) = (10/32)(9/31)(8/30)(7/29),$$ but if items are selected with replacement, it is $$P(\text{all 4 select} A) = (10/32)^4.$$
My guess it is sampling 'without replacement', but you need to make it clear.
From there, you should be able to finish the problem.
If my guesses as to the interpretation of this problem are not what you have in mind, then please say what you $do$ have in mind. Either way, maybe you should edit your problem so it is more specific.