There are four friends – Adam, Bella, Christopher and Drew. All of them are asked to choose any number in their mind. Now what is the probability that every one of them has the same number in mind? The chosen number must be in between 1 to 5. Now the answer is 101/125.. which is way off what I find.. I would like to know where I go wrong with my reasoning?
I said let A, B,C,D represent the 4 people. Let Ai denote the event that A has number i in mind and so on. Then P(each have the same number)= $P((A1,B1,C1,D1) U (A2,B2,C2,D2) U (A3,B3,C3,D3) U (A4,B4,C4,D4) U (A5,B5,C5,D5)).$ Now each Ai,Bi,Ci,Di is disjoint from the event Aj,Bj,Cj,Dj, hence the above probability is just the sum of probabilities.. So we get P(Each have the same number)$= \sum_{i=1}^{5}P(Ai,Bi,Ci,Di)= \sum_{i=1}^{5} P(Ai|Bi,Ci,Di)P(Bi|Ci,Di)P(Ci|Di)P(Di)= \sum_{i=1}^{5} P(Ai)P(Bi)P(Ci)P(Di)= 5*(1/5)^4= 1/125.$
This is way off the answer which is 101/125... Where am I going wrong? Please help!
You are not wrong. The answer is not $101/125$.
The probability every one of the other three is mindful of the same number as Adam is $1/5^3$
(Assuming independence and unbiased selections, of course.)
That is the solution to: "What is the probability that some of them are mindful of the same number?" A different problem entirely.
"All of them do it" is not the complement of "None dome of them do it".