Answer:
a)
![(7)/(15)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5g3j8124cga1xe5w3g7wocamfh3vc2es4y.png)
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
Explanation:
a) Sub of the probabilities is always 1.
Therefore,
p(red) + p(blue) + p(yellow) = 1
![(1)/(3) +(1)/(5) +p(yellow) = 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/10m8w8tworq7bd4huu3z29abtjpoajuwuy.png)
![p(yellow) = 1-(1)/(5) -(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jjo35fxcifuw65yia0eo65456ddby7kbd3.png)
![= 1-(8)/(15)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t13cm1m9sco32oz4kdiyor3paobtkolwsv.png)
![=(7)/(15)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cs6936vjmbuq4buyui5qnq06c0m1195rql.png)
b) p(red ball) =
![(n(E))/(n(S)) =(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qjm9c8c5irptkkkatdo246w3pxj7g0ughp.png)
But, it is given that n(E) = 10
Therefore,
![(10)/(n(S)) =(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o42wpzje4i5q4kz0rq7y080ocxjc0qajht.png)
n(S) = 30
Hence, there are 30 balls in total
c) From b), there are 10 red balls.
Now, p(blue ball) =
![(n(E))/(n(S)) =(1)/(5)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/e5ypp2td3bqfq35ggq9tz7tm5allu2s3f9.png)
Since, n(S) = 30,
![(n(E))/(30) =(1)/(5)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tc5klaymrlwloahjybvxii5vb1qc7xnii7.png)
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) =
![(2)/(5)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t0uv673spbc3c31ajeld1inwaddg10eja3.png)
![n(E) = (2)/(5) (40)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/260ey3arthu1y4yoro6qr2snzs22d4y383.png)
= 16
But, number of blue balls before addition = 6.
Hence, all the added balls are blue.