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  • An unbiased coin is tossed 5 times.Suppose that a variable X is assigned the value k when consecutive heads are obtained for k = 3,4,5, otherwise X takes the value -1. Then the expected value of X, is : Option: 1

An unbiased coin is tossed 5 times.Suppose that a variable X is assigned the value k when consecutive heads are obtained for k = 3,4,5, otherwise X takes the value -1. Then the expected value of X, is :
Option: 1 \frac{1}{8}


Option: 2 \frac{3}{16}


Option: 3 -\frac{1}{8}


Option: 4 -\frac{3}{16}

Answers (1)

best_answer

An unbiased coin is tossed 5 times, so total number of outcome is 25 = 32.

Now, the probability for getting number of heads occurring simultaneously

\\\mathrm{for\;k=0}\\\mathrm{E(k)=\{T T T T T\}=1}\\\mathrm{P(k)=\frac{1}{32}}

\\\mathrm{for\;k=1}\\\mathrm{E(k)=\{ HTTTT,THTTT,TTHTT,TTTHT,TTTTH,HTHTT,HTTHT,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;HTTTH,THTHT,THTTH,TTHTH,HTHTH\} =12}\\\mathrm{P(k)=\frac{12}{32}}

\\\mathrm{for\;k=2}\\\mathrm{E(k)=\{ { HHTTT,THHTT,TTHHT,TTTHH,HHTHT,HHTTH },}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; { THHTH,HTHHT,HTTHH,THTHH,HHTHH }\} =11}\\\mathrm{P(k)=\frac{11}{32}}

\\\mathrm{for\;k=3}\\\mathrm{E(k)=\{ HHHTT,THHHT,TTHHH,HHHTH,HTHHH\} =5}\\\mathrm{P(k)=\frac{5}{32}}

\\\mathrm{for\;k=4}\\\mathrm{E(k)=\{ \mathrm{HHHHT,THHHH}\} =2}\\\mathrm{P(k)=\frac{2}{32}}

\\\mathrm{for\;k=5}\\\mathrm{E(k)=\{ \mathrm{HHHHH}\} =1}\\\mathrm{P(k)=\frac{1}{32}}

\begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{K} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{P}(\mathrm{k}) & \frac{1}{32} & \frac{12}{32} & \frac{11}{32} & \frac{5}{32} & \frac{2}{32} & \frac{1}{32} \\ \hline \end{array}

\text { For } \mathrm{k}=0,1,2 ;\mathrm{X}=-1 \text { and for } \mathrm{k}=3,4,5; \mathrm{X}=\mathrm{k} \text { . }

Now expected value is

\\\sum X \times P(k)=(-1) \times \frac{1}{32}+(-1) \times \frac{12}{32}+(-1) \times \frac{11}{32} +3 \times \frac{5}{32}+4 \times \frac{2}{32}+5 \times \frac{1}{32}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{1}{8}

Posted by

Ritika Jonwal

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