From experience

Based on the experience of an insurance company employee, it was found that an insurance payout under household insurance exceeds 25,000 CZK with a probability of 0.3. What is the probability that among the next ten insurance claims:
a) at least 5 will exceed 25,000 CZK?
b) exactly one will exceed 25,000 CZK?

Final Answer:

a =  0.1503
b =  0.1211

Step-by-step explanation:

$$\begin{gathered} p=0.3 \\ n=10 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot p^0\cdot (1-p{)}^{n-0}=(\genfrac{}{}{0ex}{}{10}{0})\cdot 0.3^0\cdot (1-0.3{)}^{10-0}=1\cdot 0.3^0\cdot 0.7^{10}\doteq 0.0282 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot p^1\cdot (1-p{)}^{n-1}=(\genfrac{}{}{0ex}{}{10}{1})\cdot 0.3^1\cdot (1-0.3{)}^{10-1}=10\cdot 0.3^1\cdot 0.7^9\doteq 0.1211 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot p^2\cdot (1-p{)}^{n-2}=(\genfrac{}{}{0ex}{}{10}{2})\cdot 0.3^2\cdot (1-0.3{)}^{10-2}=45\cdot 0.3^2\cdot 0.7^8\doteq 0.2335 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot p^3\cdot (1-p{)}^{n-3}=(\genfrac{}{}{0ex}{}{10}{3})\cdot 0.3^3\cdot (1-0.3{)}^{10-3}=120\cdot 0.3^3\cdot 0.7^7\doteq 0.2668 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{n}{4}\right)\cdot p^4\cdot (1-p{)}^{n-4}=(\genfrac{}{}{0ex}{}{10}{4})\cdot 0.3^4\cdot (1-0.3{)}^{10-4}=210\cdot 0.3^4\cdot 0.7^6\doteq 0.2001 \\ {P}_4=p_0+p_1+p_2+p_3+p_4=0.0282+0.1211+0.2335+0.2668+0.2001\doteq 0.8497 \\ a=1-P_4=1-0.8497=0.1503 \end{gathered}$$

$b=p_1=0.1211$

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