Question

Let Adenote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let Bbe the event that the next request is for help with SAS. Suppose that P(A)=.30and P(B)=.50.

a. Why is it not the case that P(A)+P(B)=1?

b. Calculate P(A’).

c. Calculate P(A\( \cup \)B).

d. Calculate P(A’\( \cap \)B’).

Short answer

a. As there are other statistical software packages available.

b.\(P\left( {A'} \right) = 0.70\)

c.\(P\left( {A \cup B} \right) = 0.80\)

d. \(P\left( {A' \cap B'} \right) = 0.20\)

Step-by-step solution

Given information

A represents the event that the next request for assistance from a statistical software consultant relates to the SPSS package.

B represents the event that the next request is for help with SAS.

The probability for event A is \(P\left( A \right) = 0.30\).

The probability for event B is \(P\left( B \right) = 0.50\).

Explain why the sum of probability is not equals to 1.

a.

The sum of the probabilities of both the event is not equal to 1 because, other than SPSS and SAS, other statistical software packages are also available.

Compute the probability

b.

The probability of the provided event is computed as,

\(\begin{aligned}P\left( {A'} \right) &= 1 - P\left( A \right)\\ &= 1 - 0.30\\ &= 0.70\end{aligned}\)

Therefore, the probability of the provided event is 0.70.

c.

The probability of the provided event is computed as,

\(\begin{aligned}P\left( {A \cup B} \right) &= P\left( A \right) + P\left( B \right)\\ &= 0.30 + 0.50\\ &= 0.80\end{aligned}\)

Since A and B are disjoint events.

Therefore, the probability of the provided event is 0.80.

d.

The probability of the provided event is computed as,

\(\begin{aligned}P\left( {A' \cap B'} \right) &= P\left( {A \cup B} \right)'\\ &= 1 - P\left( {A \cup B} \right)\\ &= 1 - 0.80\\ &= 0.20\end{aligned}\)

Therefore, the probability of the provided event is 0.20.