Probability 80856

The probability of occurrence of a certain phenomenon is the same in all trials and is equal to 0.7. Attempts are repeated until this phenomenon occurs. What is the probability that we will have to make a fifth trial?

Correct answer:

p =  0.0284

Step-by-step explanation:

$$\begin{gathered} p_1=0.7 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{2}{1}\right)\cdot p_1^1\cdot (1-p_1{)}^{2-1}=(\genfrac{}{}{0ex}{}{2}{1})\cdot 0.7^1\cdot (1-0.7{)}^{2-1}=2\cdot 0.7^1\cdot 0.3^1=\frac{21}{50}=0.42 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot p_1^1\cdot (1-p_1{)}^{3-1}=(\genfrac{}{}{0ex}{}{3}{1})\cdot 0.7^1\cdot (1-0.7{)}^{3-1}=3\cdot 0.7^1\cdot 0.3^2=\frac{189}{1000}=0.189 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{4}{1}\right)\cdot p_1^1\cdot (1-p_1{)}^{4-1}=(\genfrac{}{}{0ex}{}{4}{1})\cdot 0.7^1\cdot (1-0.7{)}^{4-1}=4\cdot 0.7^1\cdot 0.3^3=\frac{189}{2500}=0.0756 \\ {p}_5=\left(\genfrac{}{}{0ex}{}{5}{1}\right)\cdot p_1^1\cdot (1-p_1{)}^{5-1}=(\genfrac{}{}{0ex}{}{5}{1})\cdot 0.7^1\cdot (1-0.7{)}^{5-1}=5\cdot 0.7^1\cdot 0.3^4\doteq 0.0284 \\ p=p_5=0.0284 \end{gathered}$$

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