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A coin is biased so that the head is 4 times as likely to occur as tail. If the coin is tossed thrice, find the probability distribution of number of tails. Hence find the mean and variance of the distribution.

 

 

 

 
 
 
 
 

Answers (1)

P\left ( Head \right )= 4p\left ( Tail \right )\Rightarrow p\left ( H \right )= \frac{4}{5},p\left ( T \right )= \frac{1}{5}
No.of trials  0                                 1                                   2                            3
P(X)         \left ( \frac{4}{5} \right )^{3}          3\left ( \frac{4}{5} \right )^{2}\left ( \frac{1}{5} \right )         3\left ( \frac{4}{5} \right )^{2}\left ( \frac{1}{5} \right )^{2}          \left ( \frac{1}{5} \right )^{3}
          = \frac{64}{125}              = \frac{48}{125}                       = \frac{12}{125}                      = \frac{1}{125}
No.of tails 0                               1                              2                                 3
X(p(X)):    0                             \frac{48}{125}                        \frac{24}{125}                            \frac{3}{125}
X2 P(X):   0                              \frac{48}{125}                         \frac{48}{125}                            \frac{9}{125}
Mean = \sum X^{2}P\left ( X \right )-\left [ \sum XP\left ( X \right ) \right ]^{2}=
     = \frac{105}{125}-\frac{9}{25}
     = \frac{60}{125}= \frac{12}{25}

Posted by

Ravindra Pindel

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