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 HH - heads, TT - tails.

There are 24=162^4=16 possible outcomes


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

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

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


2. The possible values of the random variable ZZ are 0,1,2,3,4.0, 1, 2, 3, 4.

Possible OutcomesZHHHH4HHHT3HHTH3HTHH3THHH3HHTT2HTHT2HTTH2THTH2THHT2TTHH2HTTT1THTT1TTHT1TTTH1TTTT0\def\arraystretch{1.5} \begin{array}{c:c} Possible \ Outcomes & Z \\ \hline HHHH & 4 \\ \hdashline HHHT & 3 \\ \hdashline HHTH & 3 \\ \hdashline HTHH & 3 \\ \hdashline THHH & 3 \\ \hdashline HHTT & 2 \\ \hdashline HTHT & 2 \\ \hdashline HTTH & 2 \\ \hdashline THTH & 2 \\ \hdashline THHT & 2 \\ \hdashline TTHH & 2 \\ \hdashline HTTT & 1 \\ \hdashline THTT & 1 \\ \hdashline TTHT & 1 \\ \hdashline TTTH & 1 \\ \hdashline TTTT & 0 \\ \hdashline \end{array}


3. We will assume that the probability of getting heads and tails is the same:


p=q=1/2p = q =1/2Possible OutcomesZHHHH4HHHT3HHTH3HTHH3THHH3HHTT2HTHT2HTTH2THTH2THHT2TTHH2HTTT1THTT1TTHT1TTTH1TTTT0\def\arraystretch{1.5} \begin{array}{c:c} Possible \ Outcomes & Z \\ \hline HHHH & 4 \\ \hdashline HHHT & 3 \\ \hdashline HHTH & 3 \\ \hdashline HTHH & 3 \\ \hdashline THHH & 3 \\ \hdashline HHTT & 2 \\ \hdashline HTHT & 2 \\ \hdashline HTTH & 2 \\ \hdashline THTH & 2 \\ \hdashline THHT & 2 \\ \hdashline TTHH & 2 \\ \hdashline HTTT & 1 \\ \hdashline THTT & 1 \\ \hdashline TTHT & 1 \\ \hdashline TTTH & 1 \\ \hdashline TTTT & 0 \\ \hdashline \end{array}


Construct the probability distribution of the random variable


z01234p(z)1/161/43/81/41/16\def\arraystretch{1.5} \begin{array}{c:c} z & 0 & 1 & 2 & 3 & 4 \\ \hline p(z) & 1/16 & 1/4 & 3/8 & 1/4 & 1/16 \end{array}

or


z01234p(z)0.06250.250.3750.250.0625\def\arraystretch{1.5} \begin{array}{c:c} z & 0 & 1 & 2 & 3 & 4 \\ \hline p(z) & 0.0625 & 0.25 & 0.375 & 0.25 & 0.0625 \end{array}




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS