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Question Number 21 by user1 last updated on 25/Jan/15
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.
$$\mathrm{A}\:\mathrm{card}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{at}\:\mathrm{random}\:\mathrm{from}\:\mathrm{a}\:\mathrm{well} \\ $$$$\mathrm{shuffled}\:\mathrm{deck}\:\mathrm{of}\:\mathrm{52}\:\mathrm{cards}.\:\mathrm{Find}\:\mathrm{the}\:\: \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{its}\:\mathrm{being}\:\mathrm{a}\:\mathrm{spade}\:\mathrm{or}\:\mathrm{a}\:\mathrm{king}. \\ $$
Answered by user1 last updated on 30/Oct/14
Let S be the sample space. n(S)=52. Let E = event of getting a spade. F = event of getting a king. E∩F= event of getting a king of spade. Clearly, n(E)=13, n(F)=4, n(E∩F)=1 P(E)=((n(E))/(n(S)))=((13)/(52))=(1/4) P(F)=(1/(13)) , P(E∩F)=(1/(52)) ∴ P(a spade or a king)=P(E∪F) = P(E)+P(F)−P(E∩F) = (1/4)+(1/(13))−(1/(52)) = (4/(13))
$$\mathrm{Let}\:{S}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sample}\:\mathrm{space}.\:{n}\left({S}\right)=\mathrm{52}.\: \\ $$$$\mathrm{Let}\:{E}\:=\:\mathrm{event}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{a}\:\mathrm{spade}. \\ $$$$\:\:{F}\:=\:\mathrm{event}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{a}\:\mathrm{king}. \\ $$$${E}\cap{F}=\:\mathrm{event}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{a}\:\mathrm{king}\:\mathrm{of}\:\mathrm{spade}. \\ $$$$\mathrm{C}{learly},\:\:\:{n}\left({E}\right)=\mathrm{13},\:{n}\left({F}\right)=\mathrm{4},\:{n}\left({E}\cap{F}\right)=\mathrm{1} \\ $$$${P}\left({E}\right)=\frac{{n}\left({E}\right)}{{n}\left({S}\right)}=\frac{\mathrm{13}}{\mathrm{52}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${P}\left({F}\right)=\frac{\mathrm{1}}{\mathrm{13}}\:\:,\:\:{P}\left({E}\cap{F}\right)=\frac{\mathrm{1}}{\mathrm{52}} \\ $$$$\therefore\:{P}\left(\mathrm{a}\:\mathrm{spade}\:\mathrm{or}\:\mathrm{a}\:\mathrm{king}\right)=\mathrm{P}\left({E}\cup{F}\right) \\ $$$$\:\:\:\:\:\:=\:{P}\left({E}\right)+{P}\left({F}\right)−{P}\left({E}\cap{F}\right) \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{13}}−\frac{\mathrm{1}}{\mathrm{52}}\:\:\:\:=\:\:\:\frac{\mathrm{4}}{\mathrm{13}} \\ $$

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