Hi
i need help with the attached question
I have attempted the question but i am stuck on parts d) to h)
My answers so far are
a) i substituted in $\displaystyle z=3$ hence my answer was 0.112
b)$\displaystyle P(X>2\mid X<3)=\frac{0.184}{0.888}$=0.207
c)$\displaystyle X\sim Bin(6,\:0.112) P(X=2)=\dbinom{6}{2}(0.112)^{2}(1-0.112)^{4}$=0.117
my guess to part d was
$\displaystyle \dbinom{6}{0}(0.112)^{0}(0.888)^{6}$=0.4903
Probability of first three not failing is $\displaystyle (0.888)^{3}$=0.7 hence answer is $\displaystyle \frac{0.4903}{0.7}$=0.7
any help is appreichated
thanks