Chapter 14, Problem 20E

a)

To determine

To find probability that two or more semester of Calculus.

Answer to Problem 20E

The probability that two or more semester of Calculus = 0.13

Explanation of Solution

Given:

P(Never) = 55% = 0.55

P(One) = 32% = 0.32s

Formula:

Addition rule:

  $P(AorB)=P(A)+P(B)$

Complement rule:

  $P(NotA)=1-P(A)$

  $\begin{array}{l}P\left(NoTwoormore\right)=P(Never)+P(One)\\ P\left(NoTwoormore\right)=0.55+0.32=0.87\end{array}$

Now,

  $\begin{array}{l}P\left(Twoormore\right)=1-P\left(NoTwoormore\right)\\ P\left(Twoormore\right)=1-0.87=0.13\end{array}$

Hence, probability that two or more semester of Calculus is 0.13

b)

To determine

To find probability that some Calculus.

Answer to Problem 20E

The probability that some Calculus = 0.45

Explanation of Solution

Given:

P(Never) = 55% = 0.55

P(One) = 32% = 0.32

P(Two or more) = 0.13

Formula:

Addition rule:

  $P(AorB)=P(A)+P(B)$

  $\begin{array}{l}P(Some)=P(One)+P\left(Twoormore\right)\\ P(Some)=0.32+0.13=0.45\end{array}$

Hence, probability that some Calculus is 0.45

c)

To determine

To find probability that no more than onesemester of Calculus.

Answer to Problem 20E

The probability that no more than one semester of Calculus = 0.32

Explanation of Solution

Given:

P(Never) = 55% = 0.55

P(One) = 32% = 0.32

The probability that no more than one semester of Calculus means only one semester of calculus.

Hence, it is 0.32