A bag contains, red, blue and yellow balls
the probability of picking a red ball is 1/3
the probability of picking a blue ball is 1/5

a.) Work out the probability of picking a yellow ball

b.) There are 10 red balls
How many balls are there in total

c.) Grace adds 10 balls to the bag so that:
P(red) = 1/4 and P(blue) = 2/5
How many of each color does she add

Please space out your answer and label 'a' 'b' and 'c'

Question

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JannetPalos JannetPalos · Answer 1 · 2019-01-26T15:29:10+01:00

Answer:

a) [tex]\frac{7}{15}[/tex]

b) There are 30 balls

c) Number of blue balls added = 10, number of red balls added = 0 and number of yellow balls added = 0

Step-by-step explanation:

a) Sub of the probabilities is always 1.

Therefore,

p(red) + p(blue) + p(yellow) = 1

[tex]\frac{1}{3} +\frac{1}{5} +p(yellow) = 1[/tex]

[tex]p(yellow) = 1-\frac{1}{5} -\frac{1}{3}[/tex]

[tex]= 1-\frac{8}{15}[/tex]

[tex]=\frac{7}{15}[/tex]

b) p(red ball) = [tex]\frac{n(E)}{n(S)} =\frac{1}{3}[/tex]

But, it is given that n(E) = 10

Therefore, [tex]\frac{10}{n(S)} =\frac{1}{3}[/tex]

n(S) = 30

Hence, there are 30 balls in total

c) From b), there are 10 red balls.

Now, p(blue ball) = [tex]\frac{n(E)}{n(S)} =\frac{1}{5}[/tex]

Since, n(S) = 30,

[tex]\frac{n(E)}{30} =\frac{1}{5}[/tex]

n(E) = 6

There are 6 blue balls and

the number of yellow balls = 30 - (10 + 6) = 14

After adding 10 balls, n(S) = 40

p(blue ball) = [tex]\frac{2}{5}[/tex]

[tex]n(E) = \frac{2}{5} (40)[/tex]

= 16

But, number of blue balls before addition = 6.

Hence, all the added balls are blue.