# Thread: [SOLVED] probability word problem help

1. ## [SOLVED] probability word problem help

A boy has 6 red and 7 blue blocks.
(i) in how many ways can he choose randomly 4 blocks.
i got 715 ways

(ii)what is the probablity that he will chose for red blocks.
i got 15/715 but i think is wrong.

(iii) what is the probability that the choice includes at least one blue block

thanks

2. i) $\displaystyle P(X=4) = \binom{13}{4}$

ii) $\displaystyle P(X=4 red) = \binom{6}{4}$ / $\displaystyle \binom{13}{4}$

iii)$\displaystyle P(X=1 blue) = \binom{7}{1}$$\displaystyle \binom{6}{3} / \displaystyle \binom{13}{4} \displaystyle P(X=2 blue) = \binom{7}{2}$$\displaystyle \binom{6}{2}$ / $\displaystyle \binom{13}{4}$

$\displaystyle P(X=3 blue) = \binom{7}{3}$$\displaystyle \binom{6}{1}$ / $\displaystyle \binom{13}{4}$

$\displaystyle P(X=4 blue) = \binom{7}{4}$ / $\displaystyle \binom{13}{4}$

3. thanks.

4. Originally Posted by llkkjj24
A boy has 6 red and 7 blue blocks.
(i) in how many ways can he choose randomly 4 blocks.
i got 715 ways

(ii)what is the probablity that he will choose four red blocks.
i got 15/715 but i think is wrong. why ?

(iii) what is the probability that the choice includes at least one blue block

thanks
If the blue blocks all look the same and the red blocks all look the same, then

(i)

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

Or....

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

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

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

$\displaystyle \color{red}\binom{6}{2}\color{blue}\binom{7}{2}=31 5=\color{red}2\ reds\color{blue}\ with\ 2\ blues$

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

total=715

(ii)

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 "the 4 are not all red".

Since all probabilities sum to 1,

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

5. 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)

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