A bag contains black, blue and yellow balls.
A ball is picked at random from the bag. The probability that the ball is blue is $\frac{5}{9}$. The probability that the ball is yellow is $\frac{2}{15}$. Find the probability that the ball is black. Answer ... Another bag contains 10 red balls, 7 white balls and 8 blue balls.
A further $n$ white balls are added to the bag. The probability of picking a white ball is now $\frac{2}{5}$. Find the value of $n$. Answer n = ...

1. Find the remaining probability for a black ball

$P(\text{black})=1-P(\text{blue})-P(\text{yellow})$
$=1-\frac{5}{9}-\frac{2}{15}$
$=\frac{45}{45}-\frac{25}{45}-\frac{6}{45}$
$=\frac{14}{45}$

1. Express the new number of white balls and the new total number of balls

After adding $n$ white balls,

Number of white balls $= 7+n$

Total number of balls $=10+7+8+n=25+n$

2. Form and solve the probability equation

$\frac{7+n}{25+n}=\frac{2}{5}$

$5(7+n)=2(25+n)$

$35+5n=50+2n$

$3n=15$

$n=5$