# Homework Help: Questions about Probability

1. Apr 7, 2013

### Skyrior

1. The problem statement, all variables and given/known data

I. A hundred seeds are planted in ten rows of ten seeds per row. Assuming that each seed independently germinates with probability 1/2, find the probability that the row with the maximum number of germination contains at least 8 seedlings.

II. Consider a randomly chosen n child family, where n>1. Let A be the event that the family has at most one boy, and B be the event that every child in the family is of the same sex. For what values of n are the events A and B independent?

III. A quadratic equation, $ax^{2}+bx+c=0$ is copied by a typist. However, the numbers standing for a, b and c are blurred and she can only see that they are integers of one digit. What is the probability that the equation she types has real roots?

IV. Two people agree to meet each other at the corner of two city streets between 1 p.m. and 2 p.m., but neither will wait for the other more than 30 minutes. If each person is equally likely to arrive at any time during the one hour period, determine the probability that they will in fact meet.

2. Relevant equations

binomial cumulative distribution frequency (GDC)
binomial probability distribution frequency (GDC)
Poisson probability and cumulative distribution frequency (GDC)
Independent events are such that P{A|B} = P(A) and P{B|A} = P(B)

$1 - (\frac{121}{128})^{10} \approx 0.430$
$n = 3$
$\frac{107}{576}$
$\frac{3}{4}$

3. The attempt at a solution

I:

I thought of using binomial cdf to calculate the rate for one row to have at least 8 germination, so:
One row = 1 - binomcdf(10, .5, 7)
Then we use this to calculate the rate that we have at least one row that fulfills this criteria:
All rows = 1 - binomcdf[10, (binomcdf(10, .5, 7), 0]
However, the answer doesn't seem to be this...

II:

I know why the answer is 3, but I do not know how to form an equation that can get you 3:

I tried doing this, but it does not seem to work:

Probability of $A = \frac{1}{2^{n-1}}$
Probability of $B = \frac{2}{n+1}$

So:

$\frac{1}{2^{n-1}} = \frac{\frac{2}{n+1} \times \frac{1}{2^{n-1}}}{\frac{2}{n+1}}$

But I have no idea how to solve this equation since it cancels out itself...

III:
because $b^{2} \geq 4ac$
So I have a very long list that turns out to be wrong:

For b = 1, 4ac is impossible to be 1 or lower
For b = 2, a and c must both be 1:
1/9 * 1/9 * 1/9
For b= 3:
1/9 * 3/81
etc.....

IV:

Sorry, I tried to understand this question, and I used:

Dividing into 60 minutes:

1:00 p.m. = 1/60 * 1/60
1:01 p.m. = 2/60 * 2/60
etc..

1:30 p.m. = 30/60 * 30/60
1:31 p.m. = 29/60 * 29/60
etc..

I added everything up, but I did not get the answer required.

Thank you for any help :)

Last edited: Apr 7, 2013
2. Apr 7, 2013

### haruspex

For I, your method looks ok, but I can't be sure without seeing the details of your working. I agree with the posted answer. (The posted answer is after a bit of cancellation. The uncancelled version is 1 - (1 - 56/1024)10.)

For II, you seem to have swapped A and B over. Your expression for A is the right one for B. For the given A (at most one boy), what is the probability of no boys? What is the probability of exactly one boy?
For the independence test, you also need to calculate the probability of the event A&B.

For III, I don't see how the answer can be the one given. If you assume no digit is zero then the denominator must be a factor of 729; if you allow any digit to be zero it must be a factor of 1000. Maybe you're supposed to assume there is no common factor - didn't try that.

The easiest way to approach IV is graphically. Draw a graph with the two arrival times as the axes. Shade the areas corresponding to their meeting.

3. Apr 8, 2013

### Skyrior

Thanks so much, for I, I checked the formula for binomcdf and it makes sense now! I also understand IV and II now! Thanks :)

4. Apr 8, 2013

### haruspex

Just figured out how you could get the denominator in III. You have to assume that none of the digits is zero, and that a and b are not 1 (because you wouldn't bother to write them if they were). But it seems to me that if you are going to make those assumptions you can justify assuming no common factors too, making the answer different again.

5. Apr 8, 2013

### Skyrior

Ah, I see. Thanks for the tip. I really appreciate it! (it also makes me wonder why the question is so ambiguous though..) :)