The probability that she gets a ride 3 days in the week is 0.309. The probability that she gets a ride to work at least 2 times in the week is 0.969.
This is binomial, as there are two outcomes, the probability of one happening does not affect the other, and there is a fixed number of trials.
For three rides to work that week,
[tex]_nC_r(p)^r\times(1-p)^{n-r}
\\
\\_5C_3(0.7)^3(1-0.7)^{5-3}=\frac{5!}{3!2!}(0.7)^3(0.3)^2=0.3087\approx0.309[/tex]
To find the probability that she gets a ride at least twice that week, we first find the probability that she does not get a ride at all that week or she only gets a ride one day; then we subtract that from 1:
[tex]1-[P(X=0)\text{ or }P(X=1)]
\\
\\=1-(_5C_0(0.7)^0(1-0.7)^{5-0}+_5C_1(0.7)^1(1-0.7)^{5-1})
\\
\\=1-(\frac{5!}{0!5!}(0.7)^0(0.3)^5+\frac{5!}{1!4!}(0.7)^1(0.3)^4)
\\
\\=1-(0.00243+0.02835)=1-0.03078=0.96922\approx0.969[/tex]
