Let "X=" the number of correctly solved questions: "X\\sim Bin(n,p)."
Given "n=10,p=0.2."
"+P(X=8)+P(X=9)+P(X=10)="
"=\\binom{10}{6}(0.2)^6(1-0.2)^{10-6}+\\binom{10}{7}(0.2)^7(1-0.2)^{10-7}+"
"+\\binom{10}{8}(0.2)^8(1-0.2)^{10-8}+\\binom{10}{9}(0.2)^9(1-0.2)^{10-9}+"
"+\\binom{10}{10}(0.2)^{10}(1-0.2)^{10-10}=210(0.2)^6(0.8)^4+"
"+120(0.2)^7(0.8)^3+45(0.2)^8(0.8)^2+"
"+10(0.2)^9(0.8)+(0.2)^{10}+="
"=0.005505024+0.000786432+0.000073728+"
"+0.000004096+0.0000001024\\approx0.0064"
The probability that he will get at least six questions correct is "0.0064."
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