Events

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

Correct result:

x₃ =  2.15 %
x₅ =  0.06 %
x₇ =  0 %

Solution:

$$\begin{gathered} p=0.84 \\ q=1-p=1-0.84={\displaystyle \frac{4}{25}}=0.16 \\ {x}_3=100\cdot q^{3-1}\cdot p=100\cdot 0.16^{3-1}\cdot 0.84={\displaystyle \frac{1344}{625}}={\displaystyle \frac{1344}{625}}\mathrm{\%}=2.15\mathrm{\%} \end{gathered}$$

${x}_5=100\cdot q^{5-1}\cdot p=100\cdot 0.16^{5-1}\cdot 0.84=0.06\mathrm{\%}$

${x}_7=100\cdot q^{7-1}\cdot p=100\cdot 0.16^{7-1}\cdot 0.84=0\mathrm{\%}$

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