We assumed in this answer that the question b is, Are the events V and M independent?
Answer:
(a). The probability that a student has either a Visa card or a MasterCard is<em> </em>
. (b). The events V and M are not independent.
Step-by-step explanation:
The key factor to solve these questions is to know that:
![\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)](https://tex.z-dn.net/?f=%20%5C%5C%20P%28V%20%5Ccup%20M%29%20%3D%20P%28V%29%20%2B%20P%28M%29%20-%20P%28V%20%5Ccap%20M%29)
We already know from the question the following probabilities:
![\\ P(V) = 0.73](https://tex.z-dn.net/?f=%20%5C%5C%20P%28V%29%20%3D%200.73)
![\\ P(M) = 0.18](https://tex.z-dn.net/?f=%20%5C%5C%20P%28M%29%20%3D%200.18)
The probability that a student has both cards is 0.03. It means that the events V AND M occur at the same time. So
The probability that a student has either a Visa card or a MasterCard
We can interpret this probability as
or the sum of both events; that is, the probability that one event occurs OR the other.
Thus, having all this information, we can conclude that
![\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)](https://tex.z-dn.net/?f=%20%5C%5C%20P%28V%20%5Ccup%20M%29%20%3D%20P%28V%29%20%2B%20P%28M%29%20-%20P%28V%20%5Ccap%20M%29)
![\\ P(V \cup M) = 0.73 + 0.18 - 0.03](https://tex.z-dn.net/?f=%20%5C%5C%20P%28V%20%5Ccup%20M%29%20%3D%200.73%20%2B%200.18%20-%200.03)
Then, <em>the probability that a student has either a Visa card </em><em>or</em><em> a MasterCard is </em>
.<em> </em>
Are the events V and M independent?
A way to solve this question is by using the concept of <em>conditional probabilities</em>.
In Probability, two events are <em>independent</em> when we conclude that
[1]
The general formula for a <em>conditional probability</em> or the probability that event A given (or assuming) the event B is as follows:
![\\ P(A|B) = \frac{P(A \cap B)}{P(B)}](https://tex.z-dn.net/?f=%20%5C%5C%20P%28A%7CB%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28B%29%7D)
If we use the previous formula to find conditional probabilities of event M given event V or vice-versa, we can conclude that
![\\ P(M|V) = \frac{P(M \cap V)}{P(V)}](https://tex.z-dn.net/?f=%20%5C%5C%20P%28M%7CV%29%20%3D%20%5Cfrac%7BP%28M%20%5Ccap%20V%29%7D%7BP%28V%29%7D)
![\\ P(M|V) = \frac{0.03}{0.73}](https://tex.z-dn.net/?f=%20%5C%5C%20P%28M%7CV%29%20%3D%20%5Cfrac%7B0.03%7D%7B0.73%7D)
![\\ P(M|V) \approx 0.041](https://tex.z-dn.net/?f=%20%5C%5C%20P%28M%7CV%29%20%5Capprox%200.041)
If M were independent from V (according to [1]), we have
Which is different from we obtained previously;
That is,
![\\ P(M|V) \approx 0.041](https://tex.z-dn.net/?f=%20%5C%5C%20P%28M%7CV%29%20%5Capprox%200.041)
So, the events V and M are not independent.
We can conclude the same if we calculate the probability
, as follows:
![\\ P(V|M) = \frac{P(V \cap M)}{P(M)}](https://tex.z-dn.net/?f=%20%5C%5C%20P%28V%7CM%29%20%3D%20%5Cfrac%7BP%28V%20%5Ccap%20M%29%7D%7BP%28M%29%7D)
![\\ P(V|M) = \frac{0.03}{0.18}](https://tex.z-dn.net/?f=%20%5C%5C%20P%28V%7CM%29%20%3D%20%5Cfrac%7B0.03%7D%7B0.18%7D)
![\\ P(V|M) = 0.1666.....\approx 0.17](https://tex.z-dn.net/?f=%20%5C%5C%20P%28V%7CM%29%20%3D%200.1666.....%5Capprox%200.17)
Which is different from
In the case that both events <em>were independent</em>.
Notice that
![\\ P(V|M)*P(M) = P(M|V)*P(V) = P(V \cap M) = P(M \cap V)](https://tex.z-dn.net/?f=%20%5C%5C%20P%28V%7CM%29%2AP%28M%29%20%3D%20P%28M%7CV%29%2AP%28V%29%20%3D%20P%28V%20%5Ccap%20M%29%20%3D%20P%28M%20%5Ccap%20V%29)
![\\ \frac{0.03}{0.18}*0.18 = \frac{0.03}{0.73}*0.73 = 0.03 = 0.03](https://tex.z-dn.net/?f=%20%5C%5C%20%5Cfrac%7B0.03%7D%7B0.18%7D%2A0.18%20%3D%20%5Cfrac%7B0.03%7D%7B0.73%7D%2A0.73%20%3D%200.03%20%3D%200.03)