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A game fair requires that you draw a queen from a deck of 52 ards to win. The cards are put back into the deck after each draw, and the deck is shuffled. That is the probability that it takes you less than four turns to win?

User Mashmagar
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1 Answer

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Step-by-step explanation

The probability (P) is winning in less than four turns can be decomposed as the following sum:

The probability of winning in one turn is


P(\text{Winning in turn 1})=(\#Queens)/(\#Cards)=(4)/(52).

The probability of winning in the second turn is


\begin{gathered} P(\text{ Winning in the second turn})=P(\text{ Lossing (in turn 1)})\cdot P(\text{ Winning (in turn 2)}), \\ \\ P(\text{ Winning in the second turn})=(\#NoQueens)/(\#Cards)\cdot(\#Queens)/(\#Cards), \\ \\ P(\text{ Winning in the second turn})=(48)/(52)\cdot(4)/(52)\text{.} \end{gathered}

The probability of winning in the third turn is


\begin{gathered} P(\text{ Winning in the third turn})=P(\text{ Lossing (in turn 1)})\cdot P(\text{ Lossing (in turn 2)})\cdot P(\text{ winning (in turn 3)}), \\ \\ P(\text{ Winning in the third turn})=(\#NoQueens)/(\#Cards)\cdot(\#NoQueens)/(\#Cards)\cdot(\#Queens)/(\#Cards), \\ \\ P(\text{ Winning in the third turn})=(48)/(52)\cdot(48)/(52)\cdot(4)/(52)\text{.} \end{gathered}

Adding all together, we get


\begin{gathered} P(\text{ Winning in less than four turns})=(4)/(52)+(48)/(52)\cdot(4)/(52)+(48)/(52)\cdot(48)/(52)\cdot(4)/(52), \\ \\ P(\text{ Winning in less than four turns})=(469)/(2197), \\ \\ P(\text{ Winning in less than four turns})\approx0.2135, \\ \\ P(\text{ Winning in less than four turns})\approx21.35\% \end{gathered}

Answer

The probability of winning in less than four turns is (approximately) 21.35%.

A game fair requires that you draw a queen from a deck of 52 ards to win. The cards-example-1
User Marshall Brekka
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