Let "X=" the number of correctly solved questions: "X\\sim Bin (n,p)."
Given "n=10,p=0.2."
"P(X\\geq 6)=P(X=6)+P(X=7)+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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