Originally Posted by

**Archie Meade** If the blue blocks all look the same and the red blocks all look the same, then

(i)

$\displaystyle \binom{13}{4}=715$

Or....

$\displaystyle \color{red}\binom{6}{4}=15=number\ of\ ways\ to\ choose\ 4\ red\ blocks$

$\displaystyle \color{blue}\binom{7}{4}=35=number\ of\ ways\ to\ choose\ 4\ blue$

$\displaystyle \color{red}\binom{6}{3}\color{blue}\binom{7}{1}=14 0=\color{red}3\ red\color{blue}\ with\ 1\ blue$

$\displaystyle \color{red}\binom{6}{2}\color{blue}\binom{7}{2}=32 5=\color{red}2\ reds\color{blue}\ with\ 2\ blues$

$\displaystyle \color{red}\binom{6}{1}\color{blue}\binom{7}{3}=21 0=\color{red}1\ red\color{blue}\ with\ 3\ blues$

total=715

(ii)

Why do you think your answer is wrong??

It was ok to be suspicious, however $\displaystyle \binom{13}{4}=715$

You get the same answer by summing the individual cases.

(iii)

the purpose of the question is for you to discover that there is a fast way to calculate (iii)

At least 1 is blue is equivalent to "all 4 are __not__ red".

Since all probabilities sum to 1,

the answer is

$\displaystyle 1-\color{red}P(all\ 4\ are\ red)\color{black}=1-\frac{\color{red}\binom{6}{4}\color{black}}{\binom {13}{4}}=1-\frac{\color{red}15}{75}$