A local bank reports that 70% of its customers maintain a ch

[Sonya]

A local bank reports that 70% of its customers maintain a chequing account, 50% have a savings account and 40% have both. If a customer is chosen at random:
a) What is the probability the customer has a chequing or savings account?
b) What is the probability the selected customer does not have either a chequing or savings account?
c) What is the probability the selected customer has a chequing account but does not have a savings account?
d) What is the probability the selected customer has exactly one of the two types of accounts?

I have no clue about this probability things and have come up with a probability of 100% for a and b which are probabily wrong. This is a correspondence course and I have contacted my tutor who by the way was absolutely no help and told me that I shouldn't worry about one unit it wouldn't fail me could you believe that I would just like to understand if someone could please break this down for me so I can understand a little more about probability thanks abunch please help me im completely lost Sonya

 

[Denis]

The probability that you should fire your tutor is 100% :shock:

 

[marcmtlca]

I will work one of them out. I would like to stress the importance of taking your time with the material and working through examples in your textbook. Anyhow, here we go.

Normally what people do is define events then use the probability laws to calculate the probabilities of said events. In this case:

Let:
C=Event that a customer has a chequing account
S=Event that a customer has a savings account

We are given that:
\(\displaystyle P(C)=\frac{7}{10}\)
\(\displaystyle P(S)=\frac{1}{2}\)
\(\displaystyle P(C \cap S)=\frac{4}{10}\)

We desire the probability that a customer has a chequing or savings account (or both is implied here i think).

This event is represented by
\(\displaystyle C \cup S\)

We then look at the probability of this event:
\(\displaystyle P(C \cup S)\)

We need to find a way of calculating this, we then use the following law:

\(\displaystyle P(C \cup S)=P(C)+P(S)-P(C \cap S)=.7+.5-.4=.8\)

For the other questions you might find the following results useful:

\(\displaystyle P(A \cap B^c)=P(A)-P(A \cap B)\)
\(\displaystyle P(A^c)=1-P(A)\)

and if A and B are disjoint, i.e. \(\displaystyle A \cap B = \emptyset\)
Then it follows that \(\displaystyle P(A \cup B)=P(A)+P(B)\). This just follows from the law I used to do a).

 

[Sonya]

probability question

thanks for the help.. I thinkI understand and can do the following questions. My textbook questions are very simple ones about dice and decks of cards and they differ alot from my actually study questions, though I very much appreciate the help