You have a 54 card deck with 2 jokers, a normal deck has 52 cards. What is the probability of removing 5 cards from the deck and then the next 2...

Sunday, December 7, 2014

You have a 54 card deck with 2 jokers, a normal deck has 52 cards. What is the probability of removing 5 cards from the deck and then the next 2...

Hello!

The events considered in this problem are not independent. Therefore we have to use the formula of conditional probability:

$\displaystyle{P}{\left({A}{B}\right)}={P}{\left({A}\right)}\cdot{P}{\left({B}{\mid}{A}\right)},$

where $\displaystyle{A}$ and $\displaystyle{B}$ are some events, $\displaystyle{A}{B}$ is the event " $\displaystyle{A}$ and $\displaystyle{B}$ " and $\displaystyle{B}{\mid}{A}$ is the event " $\displaystyle{B}$ given event $\displaystyle{A}$ ".

We'll use this formula several times for all the seven elementary events, denote them $\displaystyle{A}_{{n}}.$ The first event, $\displaystyle{A}_{{1}},$ is "the first card drawn is...

Hello!

The events considered in this problem are not independent. Therefore we have to use the formula of conditional probability:

$\displaystyle{P}{\left({A}{B}\right)}={P}{\left({A}\right)}\cdot{P}{\left({B}{\mid}{A}\right)},$

We'll use this formula several times for all the seven elementary events, denote them $\displaystyle{A}_{{n}}.$ The first event, $\displaystyle{A}_{{1}},$ is "the first card drawn is not a joker" (there are only 2 jokers, so to draw them on steps 6 and 7 all previous cards must not be a joker). Obviously $\displaystyle{P}{\left({A}_{{1}}\right)}=\frac{{52}}{{54}}.$

Then we are interested in event $\displaystyle{A}_{{2}},$ "the second card is not a joker". The conditional probability $\displaystyle{P}{\left({A}_{{2}}{\mid}{A}_{{1}}\right)}$ is also obvious: $\displaystyle{54}-{1}={53}$ cards remain and $\displaystyle{52}-{1}={51}$ are suitable, $\displaystyle{P}{\left({A}_{{2}}{\mid}{A}_{{1}}\right)}=\frac{{51}}{{53}}.$

Thus $\displaystyle{P}{\left({A}_{{1}}{A}_{{2}}\right)}=\frac{{52}}{{54}}\cdot\frac{{51}}{{53}}.$

The same way $\displaystyle{P}{\left({A}_{{1}}{A}_{{2}}{A}_{{3}}{A}_{{4}}{A}_{{5}}\right)}=\frac{{52}}{{54}}\cdot\frac{{51}}{{53}}\cdot\frac{{50}}{{52}}\cdot\frac{{49}}{{51}}\cdot\frac{{48}}{{50}}.$

The events $\displaystyle{A}_{{6}}$ and $\displaystyle{A}_{{7}}$ are "the card drawn is a joker", and they give two more factors, $\displaystyle\frac{{2}}{{49}}$ and $\displaystyle\frac{{1}}{{48}}.$

So the resulting probability is

$\displaystyle\frac{{52}}{{54}}\cdot\frac{{51}}{{53}}\cdot\frac{{50}}{{52}}\cdot\frac{{49}}{{51}}\cdot\frac{{48}}{{50}}\cdot\frac{{2}}{{49}}\cdot\frac{{1}}{{48}}.$

Some numbers are reduced and we obtain

$\displaystyle\frac{{2}}{{{54}\cdot{53}}}=\frac{{1}}{{{27}\cdot{53}}}=\frac{{1}}{{1431}}\approx{0.0007}.$ This is the answer.

2 comments:

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Sunday, December 7, 2014

You have a 54 card deck with 2 jokers, a normal deck has 52 cards. What is the probability of removing 5 cards from the deck and then the next 2...

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