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A standard 52 card deck is used. It is randomly divided into two parts of 26 cards each. A card is selected from the first set, and it is found to be the Queen of Hearts. The Queen of Hearts is then placed into the second set and a card is randomly picked from that set. What is the probability that the selected card is a Queen?

Hint: define events F0, F1, F2, F3 such that Fi corresponds to the event that there are i queens initially placed in the second set.

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What have you tried? If this is homework, then please tag it as such. –  EuYu Oct 9 '12 at 22:48
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3 Answers

There's no need to distinguish those events. The remaining three queens are uniformly distributed among $51$ cards, so the probability of each of the cards originally in the second set to be a queen is $3/51=1/17$. You have a $1$ in $27$ chance of drawing the queen of hearts and a $26$ in $27$ chance of drawing one of the other cards, whose probability to be a queen is $1$ in $17$, so the probability is

$$ \frac1{27}+\frac{26}{27}\cdot\frac1{17}=\frac{17+26}{17\cdot27}=\frac{43}{459}\approx0.09368\;. $$

As a rough check, there should be approximately $5/2$ queens in the second set now on average (not exactly because it contains slightly more than half of the unknown cards), so the probability of drawing one should be approximately $5/(2\cdot27)=5/54\approx0.09259$, which works out nicely.

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+1 Nice sanity check. –  Michael Joyce Oct 10 '12 at 2:35
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Based on your hint, can you compute $$ \mathbb{P}(Q|F_i), $$ for $i=0,1,2,3$ and where $Q$ is the event "drawing a Queen in the second pack." Then $\mathbb{P}(Q)$ is just the sum of the conditionnal probabilities abovetimes the probabilities of each $F_i$.

Edit

Just to clarify my comment, you need to be careful when making a hand of $26$ cards that has no queen, $1$ queen, etc. Normally you would divide by $\binom{52}{26}$, but knowing that you have the Queen of heart in the other pack changes things. The term in the sum would be $$ \mathbb{P}(Q|F_0)\mathbb{P}(F_0)=\frac{1}{27}\times \frac{\overbrace{\binom{48}{26}}^{26 \text{ cards that are not queens}}\times\overbrace{\binom{3}{0}}^{\text{ $0$ queen from the $3$}}}{\binom{51}{26}} $$

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Would P(F0) be = (48 C 2)/(52 C 2) ? –  user33195 Oct 9 '12 at 22:50
    
if $F_0$ is no queen in the second pack, then no. Try making a $26$ cards pack that has no queen. –  Jean-Sébastien Oct 9 '12 at 22:56
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@user33195: You can type $\Pr(F_0) = \binom{48}{2}/\binom{52}{2}$ to get it formatted nicely like $\Pr(F_0) = \binom{48}{2}/\binom{52}{2}$. –  Snowball Oct 9 '12 at 22:56
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The Queen of Hearts is definitely in the second set, and so it will be drawn with probability $1/27$. For each of the other three Queens, if it is in the second set (with probability $26/51$), then it will be drawn with probability $1/27$; otherwise it cannot be drawn. The total probability is therefore $$ P_Q = \frac{1}{27}+3\cdot\frac{26}{51}\cdot\frac{1}{27}=\frac{129}{1377}=\frac{43}{459}. $$ Note that the same method can be used to calculate the probability of drawing some other rank, say a King. For each of the four Kings, if it is in the second set (with probability $26/51$), then it will be drawn with probability $1/27$; the probability of drawing a King is then $$ P_K = 4\cdot \frac{26}{51}\cdot\frac{1}{27}=\frac{104}{1377}. $$ As a nice sanity check, note that $$ P_Q + 12P_K = \frac{129 + 12\cdot 104}{1377}=1,$$ as is should, since the twelve non-Queen ranks are all equally probable, and you must draw something when you draw from the second set.

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