I got this wrong on my exam. Can anyone show me how to solve this? A hand H of 3

Question

I got this wrong on my exam. Can anyone show me how to solve this?

A hand H of 3 random cards are dealt from an ordinary deck of 52. Let E1 denote the event that H has at least 1 Ace, E2 denote the event that H has at least 2 Aces, and let EAS denote the event that H includes the Ace of Spades. What are Pr(E1), Pr(E2) and Pr(EAS)? What is the conditional probability Pr(E2 | E1)? What is the conditional probability Pr(E2|EAS)?

Explanation / Answer

1) P(E1)= 3*(4/52)*(48/52)^2+3*(4/52)^2*(48/52)+(4/52)^3= 0.2134 P(E2)= 3*(4/52)^2*(48/52)+(4/52)^3 =0.0168 P(Eas)= 3*(1/52)*(51/52)^2=0.0554 2) P(E2/E1)= 1, P(E2 intersection E1)= P(E2), => P(E2/E1)=P(E2)/P(E2)=1 3) P(E2/Eas)= P(E2)/P(E2 intersection Eas)= 0.0168/[2*(3/52)*(48/52)^2+(3/52)^2*(48/52)]=0.153

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