Thread: question regarding binomial and exponential distributions

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

Hello,

I haven't looked at the other questions yet, but the first one is false

a) i substituted in z=3 hence my answer was 0.112

You're working with a probability density function. The probability of getting a value equal to 3, while working with a rv with pdf, is 0.
What you're asked is the probability that the man fails to do it in 3 minutes. Which means it's the probability that it takes more than 3 minutes to do the task.

So you're looking for P(Z>3), which is $\displaystyle \int_3^\infty f_Z(z) ~dz$, and not $\displaystyle f_Z(3)$

I guess you made the same mistake for question b)

is the answers for part a) $\displaystyle 1-[1-e^{-(0.5).(3)}]$=0.223
part b) $\displaystyle P(X>2\mid X<3)=\frac{0.368}{0.778}$=0.473

part c) $\displaystyle X\sim Bin(6,\:0.223) P(X=2)=\dbinom{6}{2}(0.223)^{2}(1-0.223)^{4}$=0.272