If there is a deck of cards containing 2 red cards, 3 blue cards, 4 green cards, and 1 yellow card, what is the probability of choosing a red card,...

Friday, July 6, 2012

If there is a deck of cards containing 2 red cards, 3 blue cards, 4 green cards, and 1 yellow card, what is the probability of choosing a red card,...

Hello!

You should multiply these probabilities (you computed them correctly). So the answer is $\displaystyle\frac{{2}}{{10}}\cdot\frac{{3}}{{9}}=\frac{{1}}{{5}}\cdot\frac{{1}}{{3}}=\frac{{1}}{{15}}.$

This comes from the formula of the conditional probability,

$\displaystyle{P}{\left({A}{\mid}{B}\right)}=\frac{{{P}{\left({A}{\quad\text{and}\quad}{B}\right)}}}{{{P}{\left({B}\right)}}},$

where $\displaystyle{A}{\mid}{B}$ is the event "A given B", which means event A after it is known that the event B has occurred.

In our problem, B is the event "the first card is red" and A is "the second card is blue". Then $\displaystyle{\left({A}{\quad\text{and}\quad}{B}\right)}$ means "the first card is red AND the second card is blue", $\displaystyle{\left({A}{\mid}{B}\right)}$ means "the second card will be blue after it is known that the first card was red".

Obviously $\displaystyle{P}{\left({B}\right)}=\frac{{2}}{{10}}=\frac{{1}}{{5}}$ and $\displaystyle{P}{\left({A}{\mid}{B}\right)}=\frac{{3}}{{9}}=\frac{{1}}{{3}},$ so $\displaystyle{P}{\left({A}{\quad\text{and}\quad}{B}\right)}={P}{\left({A}{\mid}{B}\right)}\cdot{P}{\left({B}\right)}=\frac{{1}}{{15}}.$

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Friday, July 6, 2012