ive square root of both sides: $$ \frac{6}{6+y}=\frac{3}{8} $$ 5. Cross-multiply and solve: $$ \begin{aligned} 48&=18+3y\\ 30&=3y\\ y&=10 \end{aligned} $$ When the same event happens twice with replacement, you may be able to take the square root. Use the positive root because probabilities and numbers of counters cannot be negative. ## Without replacement: reduce the count and the total “Without replacement” means the first counter is not put back, so the total number of counters decreases by 1 before the second pick. If the first counter was blue, then both the number of blue counters and the total number of counters go down by 1. A bag contains 3 red counters and $x$ blue counters. Two counters are taken without replacement. The probability that both are blue is $\frac{5}{14}$. Find $x$. 1. The total number of counters at the start is $x+3$. 2. The probability the first counter is blue is: $$ \frac{x}{x+3} $$ 3. If one blue counter is removed, there are $x-1$ blue counters left out of $x+2$ counters. 4. Set up the equation: $$ \frac{x}{x+3}\times \frac{x-1}{x+2}=\frac{5}{14} $$ 5. Clear the fractions by cross-multiplying: $$ 14x(x-1)=5(x+3)(x+2) $$ 6. Expand and collect terms: $$ \begin{aligned} 14x^2-14x&=5x^2+25x+30\\ 9x^2-39x-30&=0\\ 3x^2-13x-10&=0 \end{aligned} $$ 7. Factorise: $$ (3x+2)(x-5)=0 $$ 8. The solutions are $x=5$ or $x=-\frac{2}{3}$. A number of counters must be a whole number, so $x=5$. For “without replacement”, the second fraction must change. If the first counter was blue, the second blue probability uses $x-1$ blue counters and one fewer counter in total. ## Ratios: use a multiplier A ratio tells you the relative sizes of the groups, not necessarily the exact numbers. If red:blue = 3:2, write red counters as $3k$ and blue counters as $2k$. The total number of counters is then $5k$. This turns a ratio question into an algebra question. The ratio of red counters to blue counters is 3:2. Two counters are taken without replacement. The probability that both counters are blue is $\frac{14}{95}$. Find the total number of counters. 1. Let the number of red counters be $3k$ and the number of blue counters be $2k$. 2. The total number of counters is $5k$. 3. The probability the first counter is blue is: $$ \frac{2k}{5k}=\frac{2}{5} $$ 4. If one blue counter is removed, the second blue probability is: $$ \frac{2k-1}{5k-1} $$ 5. Set up the equation: $$ \frac{2}{5}\times \frac{2k-1}{5k-1}=\frac{14}{95} $$ 6. Cross-multiply and solve: $$ \begin{aligned} 95\cdot 2(2k-1)&=14\cdot 5(5k-1)\\ 190(2k-1)&=70(5k-1)\\ 380k-190&=350k-70\\ 30k&=120\\ k&=4 \end{aligned} $$ 7. The total number of counters is $5k$, so the total is 20. ## “Show that” quadratic questions Sometimes the question asks you to show a particular quadratic equation first, then solve it. Your job is to start from the probability statement and work forward until the given equation appears. There are $n$ counters in a bag. 6 of them are red and the rest are blue. Two counters are taken without replacement. The probability that both counters are blue is $\frac{1}{2}$. Show that $n^2-25n+84=0$, then find $n$. 1. Since there are 6 red counters, the number of blue counters is $n-6$. 2. The probability the first counter is blue is: $$ \frac{n-6}{n} $$ 3. If one blue counter is removed, the second blue probability is: $$ \frac{n-7}{n-1} $$ 4. Set up the probability equation: $$