document.write( "Question 329834: I Need Help please :-(!!!!
\n" ); document.write( "What is the probability that at least 3 red cards will be drawn from a 52 card deck? \n" ); document.write( "
Algebra.Com's Answer #236962 by jrfrunner(365)\"\" \"About 
You can put this solution on YOUR website!
It depends on a couple of things\r
\n" ); document.write( "\n" ); document.write( "1. How many cards are you selecting, for which you want to know P(at least 3 red)\r
\n" ); document.write( "\n" ); document.write( "2.with replacement or without replacement\r
\n" ); document.write( "\n" ); document.write( "You are asking about selecting 4 cards without replacement and probability of selecting X=at least 3 red cards\r
\n" ); document.write( "\n" ); document.write( "There are 26 red cards in a deck of 52 cards\r
\n" ); document.write( "\n" ); document.write( "choosing 3 red out of 4 cards =RRRB but there are 4C3 ways to choose that black one
\n" ); document.write( "Choosing 4 reds out of 4 cards = RRRR\r
\n" ); document.write( "\n" ); document.write( "P(X>=3)=P(X=3)+P(X=4)= 4C3*P(RRRB)+4C4*P(RRRR)=4C3*(26/52)*(25/51)*(24/50)*(26/49) + 4C4*(26/52)*(25/51)*(24/50)*(23/49)
\n" ); document.write( "= 0.2497 + 0.0552=0.30492\r
\n" ); document.write( "\n" ); document.write( "another way to solve this is
\n" ); document.write( "P(X>=3)=P(X=3)+P(X=4)= (26C3*26C1)+26C4*26C0)/52C4 \r
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