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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:

  1. Did I calculate these probabilities correctly?
  2. Are events L and R, B and N mutually exclusive events?
  3. 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?
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2 Answers 2

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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.

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  • $\begingroup$ "does not equal" implies that they are not independent here $\endgroup$
    – Henry
    Sep 4, 2022 at 1:31
  • $\begingroup$ @Henry, if I change the P[L] to be 0.5 instead of 0.6, , then P[N] * P[L] = 0.35, which equals to P[NL], does it then make L and N independent? $\endgroup$
    – user97662
    Sep 4, 2022 at 3:19
  • $\begingroup$ Yes - if you changed the "table with original data" to have $0.5$ instead of $0.6$ and left the $0.7$ and $0.35$ unchanged then $N$ and $L$ would be independent (as would $B$ and $L$, and $N$ and $R$, and $B$ and $R$) $\endgroup$
    – Henry
    Sep 4, 2022 at 3:22
  • $\begingroup$ @Henry, that is very interesting. Two (or more) events could be made independent or dependent of each other just by empirical observation, i.e., one day, N and L can be dependent, and the next day, N and L can become independent, only to rely on the customers' behavior of that day. $\endgroup$
    – user97662
    Sep 4, 2022 at 4:08
  • $\begingroup$ In the question these are random variables, so can be dependent or independent; they can be mutually exclusive and if they are then will be dependent (unless one of them has probability $0$ or $1$). Observations are different: if you do not know the probabilities then the observations may be suggestive of whether the probabilities are independent or not (for example you could use a chi-squared test) but cannot tell you for sure $\endgroup$
    – Henry
    Sep 4, 2022 at 12:22
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I think you have everything correct.

     L      R
---------------------
B   0.25   0.05   0.3
---------------------
N   0.35   0.35   0.7
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    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 $P[N]\, P[L] = 0.7 \times 0.6=0.42$.

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