A bag contains 4 black coins, 7 red coins and 11 white coins. Two coins are drawn from the bag at random, one after another without replacement.

1. Find the probability that the second coin drawn is white

Total number of coins $=4+7+11=22$.

There are $11$ white coins, so
$P(\text{white on second draw})=\frac{11}{22}=\frac{1}{2}.$

1. Find the new total number of coins after adding yellow coins

Originally, total number of coins
$=4+7+11=22$.

After adding $x$ yellow coins, total number of coins
$=22+x$.

2. Form the probability of drawing two black coins without replacement

There are still $4$ black coins.

Probability of first black coin $=\dfrac{4}{22+x}$.

After one black coin is drawn, there are $3$ black coins left out of $21+x$ coins.

So,
$\frac{4}{22+x}\times \frac{3}{21+x}=\frac{1}{50}.$

3. Solve the equation for $x$

$\frac{12}{(22+x)(21+x)}=\frac{1}{50}$
$(22+x)(21+x)=600$

Since $24\times25=600$,
$21+x=24 \quad \text{and} \quad 22+x=25.$

Hence,
$x=3.$