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# A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

$\begin{array}{1 1} \begin{matrix} \textbf{X} & 0 &1 &2 \\ \textbf{P(X)} &\large \frac{9}{16} &\large \frac{3}{8} &\large\frac{1}{16}\end{matrix} \\ \begin{matrix} \textbf{X} & 0 &1 &2 \\ \textbf{P(X)} &\large \frac{9}{16} &\large \frac{5}{8} &\large\frac{1}{16}\end{matrix} \\ \begin{matrix} \textbf{X} & 0 &1 &2 \\ \textbf{P(X)} &\large \frac{1}{16} &\large \frac{3}{8} &\large\frac{9}{16}\end{matrix} \\ \begin{matrix} \textbf{X} & 0 &1 &2 \\ \textbf{P(X)} &\large \frac{1}{16} &\large \frac{5}{8} &\large\frac{1}{16}\end{matrix}\end{array}$

Can you answer this question?

Given a biased coin such that heads is 3 times as likely as tails. The coin is tossed twice.
P(H) = $\large\frac{3}{4}$ and P(T) = $\large\frac{1}{4}$
Let X be the random variable for the number of tails.
P (X = 0) = P (HH) = $\large\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$
P (X = 1) = P (HT, TH) = $\large \frac{3}{4} \times \frac{1}{4} + \frac{3}{4} \times \frac{1}{4} = \frac{3}{8}$
P (X = 2) = P (TT) = $\large\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$
Therefore the probability distribution is as follows: $\begin{matrix} \textbf{X} & 0 &1 &2 \\ \textbf{P(X)} &\large \frac{9}{16} &\large \frac{3}{8} &\large\frac{1}{16}\end{matrix}$
answered Jun 20, 2013