Events

Event P has probability of 0.84. What is the probability that the event P occurs in 3, 5, 7 try.

Result

x3 =  2.15 %
x5 =  0.06 %
x7 =  0 %

Solution:

p=0.84 q=1p=10.84=425=0.16   x3=100 q31 p=100 0.1631 0.84=1344625=2.1504=2.15%p = 0.84 \ \\ q = 1 - p = 1 - 0.84 = \dfrac{ 4 }{ 25 } = 0.16 \ \\ \ \\ \ \\ x_{ 3 } = 100 \cdot \ q^{ 3-1 } \cdot \ p = 100 \cdot \ 0.16^{ 3-1 } \cdot \ 0.84 = \dfrac{ 1344 }{ 625 } = 2.1504 = 2.15 \%
 x5=100 q51 p=100 0.1651 0.840.0551=0.06% \ \\ x_{ 5 } = 100 \cdot \ q^{ 5-1 } \cdot \ p = 100 \cdot \ 0.16^{ 5-1 } \cdot \ 0.84 \doteq 0.0551 = 0.06 \%
 x7=100 q71 p=100 0.1671 0.840.0014=0% \ \\ x_{ 7 } = 100 \cdot \ q^{ 7-1 } \cdot \ p = 100 \cdot \ 0.16^{ 7-1 } \cdot \ 0.84 \doteq 0.0014 = 0 \%



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