Suppose a non bias coin was tossed two times?

**jim1174** · December 4th 2007, 10:54 PM

how would you do a random variable problem like this

suppose a non bias coin was tossed two times

**DivideBy0** · December 4th 2007, 11:06 PM

Originally Posted by jim1174

how would you do a random variable problem like this

suppose a non bias coin was tossed two times

Well... let $x$ be the random variable. You could make it stand for the number of times heads appeared.

$Pr(X=x)$ would then be the probability that x heads appeared.

$Pr(X=0) = 1-\left(\frac{1}{2}\right)^2 = \frac{3}{4}$ xxx
$Pr(X=1)=\frac{1}{2}$
$Pr(X=2) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Since the coin is non-biased these same probabilities would appear if you set $x$ as the number of tails that appeared.

Is this what you wanted?

**TwistedOne151** · December 5th 2007, 12:06 AM

DividyBy0,

Shouldn't it be $Pr(X=0) = \left(1-\frac{1}{2}\right)^2 = \frac{1}{4}$ , not $Pr(X=0) = 1-\left(\frac{1}{2}\right)^2 = \frac{3}{4}$ ?

--Kevin C.

**DivideBy0** · December 5th 2007, 12:25 AM

yeah, you're right

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