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Chuck has some cards in his pocket. In his pocket he has three hearts, one diamond and eight clubs. If Chuck randomly takes two cards out of his pocket, with replacement, what is the probability that he did not choose a club either time?

User Lakuzz
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\begin{gathered} \text{Given} \\ 3\text{ hearts} \\ 1\text{ diamond} \\ 8\text{ clubs} \end{gathered}
\begin{gathered} P(\text{Not Clubs})=\frac{\text{\# of hearts}+\text{\# of diamonds}}{\text{Total \# of cards}} \\ P(\text{Not Clubs})=(3+1)/(3+1+8) \\ P(\text{Not Clubs})=(4)/(12) \\ P(\text{Not Clubs})=(1)/(4) \\ \\ P(\text{Not Clubs twice})=P(\text{Not Clubs})\cdot P(\text{Not Clubs}) \\ P(\text{Not Clubs twice})=(1)/(4)\cdot(1)/(4) \\ P(\text{Not Clubs twice})=(1)/(16) \end{gathered}

Therefore, the probability of not choosing a club either time is 1/16.

User Shaheer Akram
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