A-card-is-drawn-at-random-from-a-well-shuffled-deck-of-52-cards-Find-the-probability-of-its-being-a-spade-or-a-king-

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Question Number 21 by user1 last updated on 25/Jan/15

$$\mathrm{A}\mathrm{card}\mathrm{is}\mathrm{drawn}\mathrm{at}\mathrm{random}\mathrm{from}\mathrm{a}\mathrm{well}$$$$\mathrm{shuffled}\mathrm{deck}\mathrm{of}52\mathrm{cards}.\mathrm{Find}\mathrm{the}$$$$\mathrm{probability}\mathrm{of}\mathrm{its}\mathrm{being}\mathrm{a}\mathrm{spade}\mathrm{or}\mathrm{a}\mathrm{king}.$$

Answered by user1 last updated on 30/Oct/14

$$\mathrm{Let}S\mathrm{be}\mathrm{the}\mathrm{sample}\mathrm{space}.n\left(S\right)=52.$$$$\mathrm{Let}E=\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{spade}.$$$$F=\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{king}.$$$$E\cap F=\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{king}\mathrm{of}\mathrm{spade}.$$$$\mathrm{C}learly,n\left(E\right)=13,n\left(F\right)=4,n(E\cap F)=1$$$$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{13}{52}=\frac{1}{4}$$$$P\left(F\right)=\frac{1}{13},P(E\cap F)=\frac{1}{52}$$$$\therefore P\left(\mathrm{a}\mathrm{spade}\mathrm{or}\mathrm{a}\mathrm{king}\right)=\mathrm{P}(E\cup F)$$$$=P\left(E\right)+P\left(F\right)-P(E\cap F)$$$$=\frac{1}{4}+\frac{1}{13}-\frac{1}{52}=\frac{4}{13}$$

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