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ssw513
04-09-2008, 05:55 PM
I need help! I know how to solve a basic Probability....like tossing a head on a bent coin has a given probability of 1/3 so the prob. that of tossing all heads in four tosses would be: 1H^4T^0 = 1(1/3)^4 = 1/81. But what about this one: Same probability for tossing heads on a bent coin (1/3) - in four tosses - what is the probability that AT LEAST TWO heads will be tossed (do you have to consider 1H^4T^0 and 4H^3T^1 and 6H^2T^2

pka
04-09-2008, 06:25 PM
\sum\limits_{k = 2}^4 {{\binom {4}{k}}\left( {\frac{1}{3}} \right)^k \left( {\frac{2}{3}} \right)^{4 - k} }

soroban
04-12-2008, 09:47 AM
Hello, ssw513!

\text{Tossing a head on a bent coin has a given probability of }\frac{1}{3}
\text{so the prob. that of tossing four heads would be: }\:1H^4T^0 \:= \:1\left(\frac{1}{3}\right)^4 \:= \:\frac{1}{81}