Question #303078

 Four coins are tossed. Let Z be the random variable representing the number heads that occur. Find the values of the random variable Z. ( Take note that you should have 2 = 16 possible outcomes ).


1
Expert's answer
2022-03-01T10:42:19-0500

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\}

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

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}





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