91Ó°ÊÓ

$Table3.22$identifies a group of children by one of four hair colors, and by type of hair.

Completing the $Table3.22$

the probability that a randomly selected child will have wavy hair:

$$\begin{gathered} P(childhavingWavyhair)=\frac{Totalnumberofchildrenshavewavyhair}{Totalnumberofchildrenshavealltypeofhair} \\ P(childhavingWavyhair)=\frac{43}{215} \end{gathered}$$

the probability that a randomly selected child will have either brown or blond hair:

$P(brownhairchildrens)=\frac{100}{215}$

$P(blondhairchildrens)=\frac{20}{215}$

role="math" localid="1650308122865" $$\begin{gathered} P(brownhairORblondhair)=P(brownhair)+P(blondhair)-P(brownhairANDblondhair) \\ P(brownhairORblondhair)=\frac{100}{215}+\frac{20}{215}-\left(\frac{100}{215}×\frac{20}{215}\right) \\ P(brownhairORblondhair)=\frac{120}{215}-\left(\frac{2000}{{215}^2}\right) \\ P(brownhairORblondhair)=\frac{25800-2000}{46225}=\frac{23800}{46225}=\frac{110}{215} \end{gathered}$$

the probability that a randomly selected child will have wavy brown hair:

$P(Wavybrownhair)=\frac{20}{215}$

the probability that a randomly selected child will have red hair, given that he or she has straight hair:

$P(\overline{)redhair}straighthair)=\frac{12}{172}$

B is the event of a child having brown hair, find the probability of the complement of B:

$$\begin{gathered} P\left(brow{n}^{\text{'}}\right)=1-P\left(brown\right) \\ P\left(brow{n}^{\text{'}}\right)=1-\frac{100}{215}=\frac{115}{215} \end{gathered}$$

the complement of $B$ represent Red, blond and black hair. Depends on what $"B"$ means