Probability: Question 8

Worked solution

Setting up the tree diagram

The bag starts with $5$ red and $3$ blue balls ($8$ total). Because the balls are not replaced, the probabilities on the second draw depend on the outcome of the first. The branch probabilities are:

(After the first ball is removed, $7$ balls remain, and either the red count or the blue count is reduced by $1$ depending on which colour was drawn first.)

Part (a): Both balls red

Multiplying along the “red, then red” branch: $P(\text{both red}) = \frac{5}{8}\times\frac{4}{7} = \frac{20}{56} = \frac{5}{14} \approx 0.357$

Part (b): Exactly one red ball

“Exactly one red” happens via either the “red then blue” branch or the “blue then red” branch, so their probabilities are added: $P(\text{exactly one red}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28} \approx 0.536$

Check: all four branch probabilities should sum to $1$: $\frac{20}{56}+\frac{15}{56}+\frac{15}{56}+\frac{6}{56} = \frac{56}{56} = 1$, as required.

Part (c): Conditional probability given exactly one red

Restrict attention to the event “exactly one red”, which consists of the two branches “red then blue” ($\frac{15}{56}$) and “blue then red” ($\frac{15}{56}$): $P(\text{red first} \mid \text{exactly one red}) = \frac{P(\text{red, then blue})}{P(\text{exactly one red})} = \frac{15/56}{30/56} = \frac{15}{30} = \frac{1}{2}$

Check: since $P(\text{red then blue}) = \frac{15}{56}$ is exactly equal to $P(\text{blue then red}) = \frac{15}{56}$, the two orders within “exactly one red” are equally likely, so it makes sense that, given exactly one red was drawn, there is a $\frac{1}{2}$ chance it came first, agreeing with the calculation above.

Final answers

• (a) $P(\text{both red}) = \boxed{\dfrac{5}{14}} \approx 0.357$
• (b) $P(\text{exactly one red}) = \boxed{\dfrac{15}{28}} \approx 0.536$
• (c) $P(\text{red first} \mid \text{exactly one red}) = \boxed{\dfrac{1}{2}}$