A) 0.2048
B) 0.7373
To find the probability that she is correct exactly 3 days, we have
[tex]_5C_3(0.8)^3(1-0.8)^{5-3}
\\
\\\frac{5!}{3!2!}(0.8)^3(0.2)^2=0.2048[/tex]
To find the probability that she is right at least 4 of the 5 days, we find the probability she is right 4 days, the probability she is right 5 days, and add them together:
[tex]_5C_4(0.8)^4(1-0.8)^{5-4}+_5C_5(0.8)^5(1-0.8)^{5-5}
\\
\\\frac{5!}{4!1!}(0.8)^4(0.2)^1+\frac{5!}{5!0!}(0.8)^5(0.2)^0
\\
\\5(0.8)^4(0.2)+0.8^5=0.4096+0.32768=0.73728\approx0.7373[/tex]
