2
$\begingroup$

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?
Improve this question
$\endgroup$

2 Answers 2

Reset to default
0
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ "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
0
$\begingroup$

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 $P[N]\, P[L] = 0.7 \times 0.6=0.42$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.