# Math Probability: Question Help ME Please!!?

http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/300...

and teach me how to do the questions you see there. I am stuck and don't know how to start.

Please SHOW steps & FINAL answer. Best ANSWER will be given out to anyone who helps me.

Relevance
• 8 years ago

a)

If you ask 3 people, a yes/no question of any kind,

what are the possible numbers of "yes" responses ?

What's the fewest, if they all answer "no"?

What's the most, if they all answer "yes"?

b)

Write down all possible sequences of 3 letters

consisting of only F and D.

Here's half of them:

D DD

D DF

D FD

D FF

I bet you can figure out the other four from there.

Then since Probability(F) + Probability(D) = 1

(since we assume there are no Martians included)

P(D) = 1 - P(F) = 1 - .27 = .73

Then P(some sequence of D's and F's) = individual probabilities multiplied together.

P(DFD) = .73 * .27 * .73 = .143

Substitute .73 for each D, and .27 for each F and multiply.

Shortcut: all the ones with the same letters in different orders come out the same.

Now if you have, for example, the sequence DFD,

that is 2 D's and 1 F, so the count W = 1 in this case (for the 1 F)

So you can

- make an 8 row table for all the possibilities

- and compute their probabilities

- and count the F's in each sequence

- and write that down on the side

- and then add up the probabilities for each count.

And you end up with this:

.... Probability... W

DDD 0.389017 0

DDF 0.143883 1

DFD 0.143883 1

DFF 0.053217 2

FDD 0.143883 1

FDF 0.053217 2

FFD 0.053217 2

FFF 0.019683 3

W: Probability

0: 0.389017

1: 0.431649

2: 0.159651

3: 0.019683

This is called a Binomial Distribution.