question regarding binomial and exponential distributions

**cooltowns** · Jan 23rd 2010, 04:44 PM

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 $z=3$ hence my answer was 0.112

b) $P(X>2\mid X<3)=\frac{0.184}{0.888}$ =0.207

c) $X\sim Bin(6,\:0.112) P(X=2)=\dbinom{6}{2}(0.112)^{2}(1-0.112)^{4}$ =0.117

my guess to part d was
$\dbinom{6}{0}(0.112)^{0}(0.888)^{6}$ =0.4903

Probability of first three not failing is $(0.888)^{3}$ =0.7 hence answer is $\frac{0.4903}{0.7}$ =0.7

any help is appreichated

thanks

**Moo** · Jan 24th 2010, 12:36 AM

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 $\int_3^\infty f_Z(z) ~dz$ , and not $f_Z(3)$

I guess you made the same mistake for question b)

**cooltowns** · Jan 24th 2010, 02:03 AM

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

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

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