PDA

View Full Version : Probability Using Binomial Theorem (non-fair coin)



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}

\text{ But what about this one:}

\text{In four tosses, what is the probability that }at\:least\:two\text{ heads will be tossed?}

\text{Do you have to consider: }\:1H^4T^0\,\text{ and }\,4H^3T^1\,\text{ and }\,6H^2T^2 . . Yes!
You can also consider the opposite: "less than 2 heads" = 1 head or 0 heads
. . and subtract from 1 (one).

\text{And you would have: }\;1 - \left(4H^1T^3 + 1H^0T^4\right)