CH-11-001 · Conditional probability without replacement

A bag contains 9 red counters and 8 blue counters. Two counters are taken at random without replacement. Given that at least one of the counters is red, find the probability that both counters are red. Show your working.

Probability Trees: Conditional Probability Without Replacement Booklet p. 60 & p. 62

Method

1) Multiply along branches for P(RR) and P(at least 1 R) = 1 − P(BB)

2) Divide P(RR) by P(at least 1 R)

Worked Example (CH-11-001)

A = red; B = blue. Do not replace the first item: the second total is one smaller, and the first colour count decreases by one.

First draw	P(first)	Second draw	P(second \| first)	Path probability
A	$\frac{9}{17}$	A	$\frac{1}{2}$	$\frac{9}{17}\times\frac{1}{2}$
A	$\frac{9}{17}$	B	$\frac{1}{2}$	$\frac{9}{17}\times\frac{1}{2}$
B	$\frac{8}{17}$	A	$\frac{9}{16}$	$\frac{8}{17}\times\frac{9}{16}$
B	$\frac{8}{17}$	B	$\frac{7}{16}$	$\frac{8}{17}\times\frac{7}{16}$

P(RR)=\frac{9}{34},\quad P(\text{at least one R})=1-\frac{7}{34}=\frac{27}{34}

Restrict to the given event:

P(RR\mid\text{at least one R})=\frac{9}{34}\div\frac{27}{34}=\frac{1}{3}

Ans =

\frac{1}{3}