Very Basic Probability Question

[amsh]

Hey all,

I was hoping somebody could confirm or reject an example in a probability text book I've been studying recently, as follows:

Two cards are drawn from a standard deck of 52. The first card drawn is replaced prior to the second one being drawn. What's the probability that at least one of the cards drawn is a spade?

I come up with 1/4 (13/52) on the first draw, 1/4 on the second draw, equaling 1/2, or 50% on aggregate that at least one is a spade.

My book says the answer is 7/16 (43.75%) and goes into some depth as to why this is the answer.

Am I wrong, or is it the book?

Any input greatly appreciated.

Al

[HalfSugar]

There are four possible outcomes:

Spade, Spade
Spade, No Spade
No Spade, Spade
No Spade, No Spade

Since you say that the first card drawn is replaced in the deck (presumably at random), every draw involves 13 spades and 52 cards.

Spade, Spade = (13/52 x 13/52) = 1/16
Spade, No Spade = (13/52 x 39/52) = 3/16
No Spade, Spade = (39/52 x 13/52) = 3/16
No Spade, No Spade = (39/52 x 39/52) = 9/16

Since the 'No Spade, No Spade' scenario does not satisfy our criteria, we only care about the other three scenarios and therefore the total odds are:

(1/16 + 3/16 + 3/16) = 7/16

The quick route is to just do [1- P(No Spade, No Spade)] = 1 - 9/16 = 7/16