Answer to Question #305360 in Statistics and Probability for Mae

Question #305360

Four coins are tossed.Let Z be the random variable representing of heads that occur.Construct a table and find the values of random variable Z.

1.Determine the sample space.Let represent the tail and it represent the head.

2.Count the number of head in each outcome in the sample space and assign this outcome.

3.Indicate the summary of the possible values of random variable.


1
Expert's answer
2022-03-03T17:44:33-0500

1. Let's denote "H" - heads, "T" - tails.

There are "2^4=16" possible outcomes


"S=\\{HHHH, HHHT, HHTH, HTHH, THHH,"

"HHTT, HTHT, HTTH, THTH, THHT, TTHH,"

"HTTT, THTT, TTHT, TTTH, TTTT\\}"


2. The possible values of the random variable "Z" are "0, 1, 2, 3, 4."

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n Possible \\ Outcomes & Z \\\\ \\hline\n HHHH & 4 \\\\\n \\hdashline\n HHHT & 3 \\\\\n \\hdashline\n HHTH & 3 \\\\\n \\hdashline\n HTHH & 3 \\\\\n \\hdashline\n THHH & 3 \\\\\n \\hdashline\n HHTT & 2 \\\\\n \\hdashline\n HTHT & 2 \\\\\n \\hdashline\n HTTH & 2 \\\\\n \\hdashline\n THTH & 2 \\\\\n \\hdashline\n THHT & 2 \\\\\n \\hdashline\n TTHH & 2 \\\\\n \\hdashline\n HTTT & 1 \\\\\n \\hdashline\n THTT & 1 \\\\\n \\hdashline\n TTHT & 1 \\\\\n \\hdashline\n TTTH & 1 \\\\\n \\hdashline\n TTTT & 0 \\\\\n \\hdashline\n\\end{array}"


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 & Z \\\\ \\hline\n HHHH & 4 \\\\\n \\hdashline\n HHHT & 3 \\\\\n \\hdashline\n HHTH & 3 \\\\\n \\hdashline\n HTHH & 3 \\\\\n \\hdashline\n THHH & 3 \\\\\n \\hdashline\n HHTT & 2 \\\\\n \\hdashline\n HTHT & 2 \\\\\n \\hdashline\n HTTH & 2 \\\\\n \\hdashline\n THTH & 2 \\\\\n \\hdashline\n THHT & 2 \\\\\n \\hdashline\n TTHH & 2 \\\\\n \\hdashline\n HTTT & 1 \\\\\n \\hdashline\n THTT & 1 \\\\\n \\hdashline\n TTHT & 1 \\\\\n \\hdashline\n TTTH & 1 \\\\\n \\hdashline\n TTTT & 0 \\\\\n \\hdashline\n\\end{array}"


Construct the probability distribution of the random variable


"\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n z & 0 & 1 & 2 & 3 & 4 \\\\ \\hline\n p(z) & 1\/16 & 1\/4 & 3\/8 & 1\/4 & 1\/16\n\\end{array}"

or


"\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n z & 0 & 1 & 2 & 3 & 4 \\\\ \\hline\n p(z) & 0.0625 & 0.25 & 0.375 & 0.25 & 0.0625\n\\end{array}"




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