Two bowls are on a teacher's desk. One contains hard candy while the other contains gum. The bowl that contains the hard candy has 10 peppermints, 5 cinnamon disks, and 7 butterscotch. The bowl that contains the gum has 3 bubble gums, 9 Dentine, and 6 spearmints. If you draw one piece from each bowl, find the following probabilities.

P(peppermints, bubblegums)_

P(not butterscotch, Dentine)_

P(cinnamon or butterscotch, spearmint)_

Respuesta :

Answer:

(a)5/66

(b)15/44

(c)2/11

Step-by-step explanation:

Candy Bowl

Number of Peppermints=10

Number of cinnamon disks=5

Number of butterscotch=7

Total=10+5+7=22

Gum Bowl

Number of bubble gums=3

Number of Dentine=9

Number of spearmints=6

Total=3+9+6=18

(a)P(peppermints, bubblegums)

P(peppermints)[tex]=\frac{10}{22}[/tex]

P(bubblegums)[tex]=\frac{3}{18}[/tex]

P(peppermints, bubblegums)[tex]=\frac{10}{22}X \frac{3}{18} =\frac{5}{66}[/tex]

(b)P(not butterscotch, Dentine)

P(butterscotch)[tex]=\frac{7}{22}[/tex]

P(not butterscotch)[tex]=1-\frac{7}{22}=\frac{15}{22}[/tex]

P(Dentine)[tex]=\frac{9}{18}[/tex]

P(not butterscotch, Dentine)[tex]=\frac{15}{22} X \frac{9}{18}=\frac{15}{44}[/tex]

(c)P(cinnamon or butterscotch, spearmint)

[tex]\text{P(cinnamon)=}\frac{5}{22} \\\text{P(butterscotch)=}\frac{7}{22} \\\text{P(cinnamon or butterscotch)=}\frac{5}{22} +\frac{7}{22} =\frac{12}{22} \\\text{P(spearmint)=}\frac{6}{18} \\\text{P(cinnamon or butterscotch, spearmint)=}\frac{12}{22} X \frac{6}{18} =\frac{2}{11}[/tex]