Question on calculating probability of independent and mutually exclusive events I am given the following question:
Monitor customer behavior in the Phonesmart store. Classify the behavior as buying (B) if a customer purchases a smartphone. Otherwise the behavior is no purchase (N). Classify the time a customer is in the store as long (L) if the customer
stays more than three minutes; otherwise classify the amount of time as rapid
(R). Based on experience with many customers, we use the probability model
P[N] = 0.7, P[L] = 0.6, P[NL] = 0.35. Find the following probabilities:
a) $P[B\cup L]$
b) $P[N\cup L]$
c) $P[N\cup B]$
d) $P[LR]$
Here is my approach, create a table listing L R B N
The table with original data:
     L      R
---------------------
B   
---------------------
N   0.35          0.7
---------------------
    0.6   

Then I just fill in the blanks:
     L      R
---------------------
B   0.25   0.05   0.3
---------------------
N   0.35   0.35   0.7
---------------------
    0.6    0.4    1.0

so then
a) $P[B\cup L] = P[B] + P[L] - P[B\cap L] = 0.3 + 0.6 - 0.25 = 0.65$
b) $P[N\cup L] = P[N] + P[L] - P[N\cap L] = 0.7 + 0.6 - 0.35 = 0.95$
c) $P[N\cup B] = P[N] + P[B] = .7 + .3 = 1$
d) $P[LR] = 0$ since L and R are mutually exclusive.
Questions:

*

*Did I calculate these probabilities correctly?

*Are events L and R,  B and N mutually exclusive events?

*Are events (B, L),  (B, R), (N, L), (N, R) independent events?
If the answer to 3 is Yes, then I am confused, according to the independence definition: $\text{Events A and B are independent if and only if }P[AB] = P[A]*P[B]$  P[NL] from the table is 0.35, but according to the definition, P[NL] = 0.7 * 0.6 = 0.42. Where did I get it wrong?

 A: To answer my own question #3.
I suppose, since P[NL] = 0.35, and does not equal to P[N] * P[L] = 0.7 * 0.6 = 0.42, event N and L are are independent events.
Knowing how long a customer stays in a store can predict the likelihood of spending money by said customer.
However, if P[L] and P[R] each has a probability of 0.5, then event N and L can become two independent events:
Assume L and R each has a probability of 50%

     L      R
---------------------
B   0.15   0.05   0.3
---------------------
N   0.35   0.35   0.7
---------------------
    0.5    0.5    1.0

then P[NL] = P[N] * P[L] = 0.7 * 0.5 = 0.35, only then the duration of a customer staying in the shop has no influence on their purchase decision.
A: I think you have everything correct.
     L      R
---------------------
B   0.25   0.05   0.3
---------------------
N   0.35   0.35   0.7
---------------------
    0.6    0.4    1.0

looks reasonable so I would use addition of the components as a check and say

*

*$P[B\cup L] = P[B\cap L] + P[B \cap R] + P[N\cap L] = 0.25 + 0.05 +0.35  = 0.65$


*$P[N\cup L] = P[N\cap L] + P[N \cap R] + P[B\cap L] = 0.35 + 0.35 +0.25  = 0.95$


*$P[N\cup B] = P[N\cap L] + P[N \cap R] + P[B\cap L] + P[B\cap R] = 0.35 + 0.35 +0.25+0.05  = 1$
As described: L and R are mutually exclusive (otherwise helps here); similarly N and B are mutually exclusive.
The events are not independent: as you say for example $P[N\cap L]=0.35$, which is not equal to $P[N]\, P[L] = 0.7 \times 0.6=0.42$.
