# Thread: Probability Problem...

1. ## Probability Problem...

$\text{suppose we need to interview 150 blind men in a city(door to door survey).}$
$\text{the probability that a house is occupied is 0.73.}$
$\text{the probability that the house, if occupied, has at least one man is 0.90.}$
$\text{ the probability , given a man occupies the home, the man is blind is 0.15.}$
$\text{the likelihood that a blind man completes the interview is 0.95.}$

$\text{how many houses do we have to interview in order to have a 75\% chance}$
$\text{that we obtain 150 complete interviews from the blind men?}$

here's what I have attempted:

$\mbox{P(one complete interview is obtained from a house with a blind man)}\;=\; 0.73(0.90)(0.15)(0.95)=0.0937$

$\text{no of houses we would need to survey to complete 150 interviews} \;=\; \frac{150}{0.0937} = 1600$

now for the last part ,

$\text{With 75\% chance to obtain 150 interviews, we need,}\;1600 \times 0.75 \;=\;1200$

is this the right approach?

2. ## Re: Probability Problem...

Originally Posted by chutiya
$\text{the probability that the house, if occupied, has at least one blind man is 0.90.}$
I'm not convinced this should say "blind".

3. ## Re: Probability Problem...

Given 1600 door knocks, you have a 50% chance of obtaining your required interviews. You need MORE thatn 1600 to get a 75% chance.

N is the number of doors needed.
Mean = N * 0.0937 = 150
Variance = N * 0.0937 * (1-0.0937)

There's enough that you probably can get away with a Normal Approximation.

What's next?

4. ## Re: Probability Problem...

Originally Posted by TKHunny
I'm not convinced this should say "blind".
Sorry. that was a typo.. it should just be $\text{P(house, if occupied, has at least one man)=0.90}$

Originally Posted by TKHunny
Given 1600 door knocks, you have a 50% chance of obtaining your required interviews. You need MORE thatn 1600 to get a 75% chance.

N is the number of doors needed.
Mean = N * 0.0937 = 150
Variance = N * 0.0937 * (1-0.0937)

There's enough that you probably can get away with a Normal Approximation.

What's next?
I do not understand why we have only a 50% chance of obtaining the required interviews with N =1600?

And why is it that with N being the number of doors needed, you have mentioned that Mean = N * 0.0937 = 150. Doesn't this mean that N =1600???

5. ## Re: Probability Problem...

Obscure hints aren't helping? I should try to be more clear.

You need a value, N, such that.

N*0.0937 = Mean > 150

N*0.0937*(1-0.0937) = Variance

SD = sqrt(Variance)

Mean - 0.67449*SD = 150

Your challenge, should you choose to accept it, is to tell me where I dreamed up that "-0.67449".

6. ## Re: Probability Problem...

0.67449 is the z value for probability 0.75, right?

So, I used $\text{Mean}- 0.67449*SD = 150$ to find N.

Thank You!!

7. ## Re: Probability Problem...

Using a Normal Approximation, that is 75% to the right and 25% to the left. Good.

8. ## Re: Probability Problem...

Finally, I understand this normal approximation thing. Thank you so much for your hints!