Tom, a university freshman, makes one to five new friends every week, with equal probability. The number of friends he makes each week is independent from all other weeks. We consider two consecutive weeks in the following questions.
- Let event A be “Tom made a total of 10 friends during the two weeks.”
- Let event B be “Tom make more than 5 friends during the two weeks.”
- Let event C be “Tom made exactly 5 friends during the first week.”
(a) Are events A and B independent? Please justify your reason.
(b) Are events A and C independent? Please justify your reason.
This is what I've come up with myself so far:
P(A) = 1/25
P(B) = 1/25 << I am not confident on this one though
P(C) = 1/5
Formulas P(B|A) = P(B) or P(A and B) = P(B ∩ A) = P(B) × P(A) To tell if events are independent or dependent
I got P(B∩A) = 1/625 and P(B|A) = 1/25
With these equations above I think that events A and B are independent and events A and C are dependent.
Any help would be appreciated. Thank you
I didn't see your question, but I assume you are here to ask if your results are correct.
$P(A) = \frac{1}{5}$ and $P(C) = \frac{1}{5}$ are right. $P(B)$ is not correct.
Denote $X$ as the number of the first week and $Y$ as the second. We have $A = \{X+Y=10\}$, $B = \{X+Y>5\}$ and $C=\{X+Y=5\}$
$P(B) = P(X+Y>5) = P(X=1,Y=5) + P(X=2,Y \ge 4), + P(X=3, Y\ge 3) + P(X=4,Y\ge2) + P(X=5) = \frac{1}{25}+\frac{2}{25}+\frac{3}{25}+\frac{4}{25}+\frac{1}{5} = \frac{3}{5}.$
So $P(B)=\frac{3}{5} \neq \frac{1}{25}$
From the "$X,Y$" representation of $A,B$ and $C$, we can see: $A\cap C = \emptyset$ and $B\cap C = \emptyset$, which means $P(A\cap C) = P(B\cap C) = 0$. Because $P(A),P(B)$ and $P(C)$ are all positive, so $(A,C)$ and $(B,C)$ are both not independant.
And we calculate $P(A\cap B) = P(\{X+Y=10\} \cap \{X+Y>5\}) = P(\{X+Y=10\}) = P(A) = \frac{1}{25}$, so $P(A\cap B) \neq P(A)$.
As a result, $P(B)$ should equal to $\frac{3}{5}$ and every two of them are not independent.