Crime solving probability

The police solved 21 crimes in the monitored period. The probability of solving a crime is 0.64. What is the probability that the police:

a) just solved 7 crimes
b) did not solve at least 2 criminal offenses

P.S. Let's assume ideal police officers

Final Answer:

p₇ =  0.0031
b =  0

Step-by-step explanation:

$$\begin{gathered} p=0.64 \\ {C}_7(21)=\left(\genfrac{}{}{0ex}{}{21}{7}\right)=\frac{21!}{7!(21-7)!}=\frac{21\cdot 20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=116280 \\ {p}_7=\left(\genfrac{}{}{0ex}{}{21}{7}\right)\cdot p^7\cdot (1-p{)}^{21-7}=116280\cdot 0.64^7\cdot (1-0.64{)}^{21-7}=0.0031 \end{gathered}$$

$$\begin{gathered} C_0(21)=\left(\genfrac{}{}{0ex}{}{21}{0}\right)=\frac{21!}{0!(21-0)!}=\frac{1}{1}=1 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{21}{0}\right)\cdot p^0\cdot (1-p{)}^{21-0}=1\cdot 0.64^0\cdot (1-0.64{)}^{21-0}=4.8123\cdot 10^{-10} \\ {C}_1(21)=\left(\genfrac{}{}{0ex}{}{21}{1}\right)=\frac{21!}{1!(21-1)!}=\frac{21}{1}=21 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{21}{1}\right)\cdot p^1\cdot (1-p{)}^{21-1}=21\cdot 0.64^1\cdot (1-0.64{)}^{21-1}=1.7966\cdot 10^{-8} \\ b=p_0+p_1=4.8123\cdot 10^{-10}+1.7966\cdot 10^{-8}=0 \end{gathered}$$

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