Answer:
P(X = 0) = [TeX]\frac{1}{16}[/TeX]
P(X = 1) = [TeX]\frac{4}{16}[/TeX]
P(X = 2) = [TeX]\frac{6}{16}[/TeX]
P(X = 3) = [TeX]\frac{4}{16}[/TeX]
P(X = 4) = [TeX]\frac{1}{16}[/TeX]
Step-by-step explanation:
The sample space for a coin tossed 4 types is given below:
[TeX]\left|\begin{array}{|c||c|c|}Occurrence & Probability & Value of Random Variable(X) \\ HHHH & \frac{1}{16} & 4\\ HHHT & \frac{1}{16} & 3\\ HHTH & \frac{1}{16} & 3\\ HTHH & \frac{1}{16} & 3\\ THHH& \frac{1}{16} & 3\\ HHTT & \frac{1}{16} & 2\\ HTHT & \frac{1}{16} & 2\\ HTTH & \frac{1}{16} & 2\\ THHT & \frac{1}{16} & 2\\ THTH & \frac{1}{16} & 2\\ TTHH & \frac{1}{16} & 2\\ HTTT & \frac{1}{16} & 1\\ THTT & \frac{1}{16} & 1\\ TTHT & \frac{1}{16} & 1\\TTTH & \frac{1}{16} & 1\\ TTTT & \frac{1}{16} & 0\\ \end{array} \right| [/TeX]
The Probability distribution for the number of heads, X is given as:
P(X = 0) = [TeX]\frac{1}{16}[/TeX]
P(X = 1) = [TeX]\frac{4}{16}[/TeX]
P(X = 2) = [TeX]\frac{6}{16}[/TeX]
P(X = 3) = [TeX]\frac{4}{16}[/TeX]
P(X = 4) = [TeX]\frac{1}{16}[/TeX]
