Respuesta :
Answer:
(a)[tex]\frac{5}{7},\frac{1}{6}[/tex] (b)[tex]\frac{1}{7},\frac{1}{6}[/tex] (c)[tex]\frac{5}{7},\frac{2}{3},\frac{3}{5}[/tex]
Step-by-step explanation:
GIVEN: Miranda has a bag of marbles with [tex]5[/tex] blue marbles, [tex]1[/tex] white marbles, and
TO FIND: a) A Blue, then a red Preview
,b)A red, then a white Preview
,c) A Blue, then a Blue, then a Blue.
SOLUTION:
Total marbles in bag [tex]=7[/tex]
(a)
Probability of drawing blue marble [tex]=\frac{\text{total blue marble}}{\text{total marbles}}[/tex]
[tex]=\frac{5}{7}[/tex]
As marble is not returned to bag,
Probability of drawing red marble [tex]=\frac{\text{total red marble}}{\text{total marbles}}[/tex]
[tex]=\frac{1}{6}[/tex]
(b)
Probability of drawing red marble [tex]=\frac{\text{total red marble}}{\text{total marbles}}[/tex]
[tex]=\frac{1}{7}[/tex]
As marble is not returned to bag,
Probability of drawing white marble [tex]=\frac{\text{total white marble}}{\text{total marbles}}[/tex]
[tex]=\frac{1}{6}[/tex]
(c)
Probability of drawing blue marble [tex]=\frac{\text{total blue marble}}{\text{total marbles}}[/tex]
[tex]=\frac{5}{7}[/tex]
As marble is not returned to bag,
Probability of drawing second blue marble [tex]=\frac{\text{total blue marble}}{\text{total marbles}}[/tex]
[tex]=\frac{4}{6}=\frac{2}{3}[/tex]
As marble is not returned to the bag
Probability of drawing third blue marble [tex]=\frac{\text{total blue marble}}{\text{total marbles}}[/tex]
[tex]=\frac{3}{5}[/tex]
