There are "2^3=8" possible outcomes
"S=\\{HHH, HHT, HTH, THH,"
"HTT, THT, TTH, TTT\\}"The possible values of the random variable "H" are "0, 1, 2, 3."
We will assume that the probability of getting heads and tails is the same:
"p = q =1\/2""\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n Possible \\ Outcomes & H \\\\ \\hline\n HHH & 3 \\\\\n \\hdashline\n HHT & 2 \\\\\n \\hdashline\n HTH & 2 \\\\\n \\hdashline\n THH & 2 \\\\\n \\hdashline\n HTT & 1 \\\\\n \\hdashline\n THT & 1 \\\\\n \\hdashline\n TTH & 1 \\\\\n \\hdashline\n TTT & 0 \\\\\n \\hdashline \n\\end{array}"
Construct the probability distribution of the random variable
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n h & 0 & 1 & 2 & 3 \\\\ \\hline\n p(h) & 1\/8 & 3\/8 & 3\/8 & 1\/8\n\\end{array}" or
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n h & 0 & 1 & 2 & 3 \\\\ \\hline\n p(h) & 0.125 & 0.375 & 0.375 & 0.125 \n\\end{array}"
Comments
Leave a comment