91Ó°ÊÓ

A student goes to the library. Let events $B=$the student checks out a book and D= the student checks out a DVD. Suppose that $P\left(B\right)=0.40,P\left(D\right)=0.30$and $P(Dâˆ\pounds B)=0.50$.

We have

$\begin{array}{r}P\left(B\right)=0.40\\ P\left(D\right)=0.30\\ P(Dâˆ\pounds B)=0.50\end{array}$

We need to calculate $P\left(B^{\text{'}}\right)$

$P\left(B^{\text{'}}\right)=1-P\left(B\right)$

Substituting the values, we get

$P\left(B^{\text{'}}\right)=1-0.40$

$=0.60$

We have

$P\left(B\right)=0.40$

$P(Dâˆ\pounds B)=0.50$

We need to calculate $P(DANDB)$

$P(\mathrm{D}ANDB)=P(Dâˆ\pounds B)×P\left(B\right)$

Substituting the values, we get

$\begin{array}{r}P(\mathrm{D}ANDB)=0.5×0.4\\ P(DANDB)=0.2\end{array}$

We have

$\begin{array}{r}P\left(B\right)=0.40\\ P\left(D\right)=0.30\\ P(DANDB)=0.2\\ P(Dâˆ\pounds B)=0.50\end{array}$

We need to calculate $P(Bâˆ\pounds D)$

role="math" localid="1648032689213" $P(Bâˆ\pounds D)=\frac{P(DANDB)}{P\left(D\right)}$

Substituting the values, we get

$\begin{array}{r}P(Bâˆ\pounds D)=\frac{0.2}{0.30}\\ P(Bâˆ\pounds D)=0.67\end{array}$

We have

$\begin{array}{r}P\left(B\right)=0.40\\ P\left(D\right)=0.30\\ P(Dâˆ\pounds B)=0.50\end{array}$

We need to calculate $P\left(DAND{B}^{\text{'}}\right)$

$P\left(DAND{B}^{\text{'}}\right)=P\left(D\right)-P(DANDB)$

Substituting the values, we get

$P\left(DAND{B}^{\text{'}}\right)=0.3-0.2$

$\begin{array}{r}P\left(DAND{B}^{\mathrm{′}}\right)=0.3−0.2\\ P\left(DAND{B}^{\mathrm{′}}\right)=0.1\end{array}$

We have

$\begin{array}{r}P\left(B\right)=0.40\\ P\left(D\right)=0.30\\ P\left(DANDB\right)=0.2\\ P(Dâˆ\pounds B)=0.50\end{array}$

We need to calculate $P\left(Dâˆ\pounds B^{\text{'}}\right)$

$P(Bâˆ\pounds D)=\frac{P(DANDB)}{P\left(D\right)}$

Substituting the values, we get

$\begin{array}{r}P(Bâˆ\pounds D)=\frac{0.2}{0.30}\\ P\left(Dâˆ\pounds B^{\mathrm{′}}\right)=0.06\end{array}$