maddoggsaddog
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- May 9, 2013
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Suppose you draw 18 cards from a standard deck of 52 cards with replacement. What is the probability of drawing at least 8 diamonds?
HELP, the impossible!
- Thread starter maddoggsaddog
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The probability would be minimal since 18/52 yields x number of diamonds out of the 13 possible. If your looking for the probability of 8 diamonds rather then 13 then x=8 out of .346 (18/52) which are given. The answer depends on how you relate those values to one another.
Hello, maddoggsaddog!
For any one draw: .\(\displaystyle \begin{Bmatrix}P(\diamondsuit) &=& \frac{13}{52} &=& \frac{1}{4} \\ P(\sim\!\diamondsuit) &=& \frac{39}{52} &=& \frac{3}{4} \end{Bmatrix}\)
We have:
. . \(\displaystyle \begin{array}{ccc} P(8\diamondsuit) &=& {18\choose8}(\frac{1}{4})^8(\frac{3}{4})^{10} \\ P(9\diamondsuit) &=& {18\choose9}(\frac{1}{4})^9(\frac {3}{4})^9 \\ P(10\diamondsuit) &=& {18\choose10}(\frac{1}{4})^{10}(\frac{3}{4})^8 \\ \vdots && \vdots \\ P(17\diamondsuit) &=& {18\choose17}(\frac{1}{4})^{17}(\frac{3}{4})^1 \\ P(18\diamondsuit) &=& {18\choose18}(\frac{1}{4})^{18}(\frac{3}{4})^0 \end{array}\)
Then add the eleven probabilities.
For any one draw: .\(\displaystyle \begin{Bmatrix}P(\diamondsuit) &=& \frac{13}{52} &=& \frac{1}{4} \\ P(\sim\!\diamondsuit) &=& \frac{39}{52} &=& \frac{3}{4} \end{Bmatrix}\)
We have:
. . \(\displaystyle \begin{array}{ccc} P(8\diamondsuit) &=& {18\choose8}(\frac{1}{4})^8(\frac{3}{4})^{10} \\ P(9\diamondsuit) &=& {18\choose9}(\frac{1}{4})^9(\frac {3}{4})^9 \\ P(10\diamondsuit) &=& {18\choose10}(\frac{1}{4})^{10}(\frac{3}{4})^8 \\ \vdots && \vdots \\ P(17\diamondsuit) &=& {18\choose17}(\frac{1}{4})^{17}(\frac{3}{4})^1 \\ P(18\diamondsuit) &=& {18\choose18}(\frac{1}{4})^{18}(\frac{3}{4})^0 \end{array}\)
Then add the eleven probabilities.
That's a special deck, Soroban! Having 18 diamonds and all
XD