Tree diagrams questions with answers

Questions

1. Question 1With replacement

   A bag contains $3$ red and $7$ blue balls. Two balls are taken at random, the first being put back before the second is taken. The tree diagram shows the outcomes. Find the probability that both balls are red.

   Show the worked solution

   1. Step 1: Read the probability on each branch of the treeFirst: $P(R) = \dfrac{3}{10}$, $P(B) = \dfrac{7}{10}$.After a red: $P(R) = \dfrac{3}{10}$, $P(B) = \dfrac{7}{10}$.After a blue: $P(R) = \dfrac{3}{10}$, $P(B) = \dfrac{7}{10}$.
   2. Step 2: Follow the path for this outcome: combine the probabilities on it$\mathstrut P(RR) = \dfrac{3}{10} \times \dfrac{3}{10}$$P(RR) = \dfrac{9}{100}$
2. Question 2Without replacement

   A bag contains $4$ red and $6$ blue balls. Two balls are taken at random, one after the other, without replacement. The tree diagram shows the outcomes. Find the probability that at least one ball is red.

   Show the worked solution

   1. Step 1: Read the probability on each branch of the treeFirst: $P(R) = \dfrac{2}{5}$, $P(B) = \dfrac{3}{5}$.After a red: $P(R) = \dfrac{1}{3}$, $P(B) = \dfrac{2}{3}$.After a blue: $P(R) = \dfrac{4}{9}$, $P(B) = \dfrac{5}{9}$.
   2. Step 2: At least one red = everything except both blue$\mathstrut P(BB) = \dfrac{3}{5} \times \dfrac{5}{9}$$P(BB) = \dfrac{1}{3}$
   3. Step 3: Take it away from 1$\mathstrut P = 1 - \dfrac{1}{3}$$P = \dfrac{2}{3}$
3. Question 3Draw your own tree

   A bag contains $5$ red and $3$ blue balls. Two balls are taken at random, the first being put back before the second is taken. Draw a tree diagram, then find the probability that both balls are red.

   Show the worked solution

   1. Step 1: Complete the tree: the first ball, then the second ball after each colour
   2. Step 2: Follow the path for this outcome: combine the probabilities on it$\mathstrut P(RR) = \dfrac{5}{8} \times \dfrac{5}{8}$$P(RR) = \dfrac{25}{64}$
4. Question 4An unknown number of balls: a quadratic

   A bag contains $n$ red balls and $3$ blue balls. Two balls are taken at random, one after the other, without replacement. The probability that both balls are red is $\dfrac{5}{14}$. Find the total number of balls in the bag.

   Show the worked solution

   1. Step 1: Multiply along the red-red branch: n red out of n + k, then one red fewer and one ball fewer$\dfrac{n}{n + 3} \times \dfrac{n - 1}{n + 2} = \dfrac{5}{14}$
   2. Step 2: Cross-multiply and expand: a quadratic equation in n$14n(n - 1) = 5(n + 3)(n + 2)$$14n^2 - 14n = 5n^2 + 25n + 30$$3n^2 - 13n - 10 = 0$
   3. Step 3: Solve: n is a number of balls, so it must be a positive whole number$(n - 5)(3n + 2) = 0$$n = 5$ or $n = -\dfrac{2}{3}$A number of balls can't be negative, so $n = 5$.
   4. Step 4: The total number of balls is n + the blue ballsTotal $= 5 + 3$$= 8$
5. Question 5Three balls: P(all three) gives a quadratic

   A bag contains $n$ red balls and $2$ blue balls. Three balls are taken at random, one after the other, without replacement. The probability that all three balls are red is $\dfrac{2}{7}$. Show that $5n^2 - 27n + 10 = 0$, and hence find the number of red balls in the bag.

   Show the worked solution

   1. Step 1: Multiply along the red-red-red branch: one red fewer and one ball fewer each time$\dfrac{n}{n + 2} \times \dfrac{n - 1}{n + 1} \times \dfrac{n - 2}{n} = \dfrac{2}{7}$
   2. Step 2: The n's cancel: cross-multiply what is left$\dfrac{(n - 1)(n - 2)}{(n + 1)(n + 2)} = \dfrac{2}{7}$$7(n - 1)(n - 2) = 2(n + 1)(n + 2)$
   3. Step 3: Expand, collect on one side and divide by the common factor: the equation to show$7n^2 - 21n + 14 = 2n^2 + 6n + 4$$5n^2 - 27n + 10 = 0$
   4. Step 4: Solve: n must be a whole number, at least 3 (three red balls are taken)$(n - 5)(5n - 2) = 0$$n = 5$ or $n = \dfrac{2}{5}$ (rejected)