If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. 

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

009.  Binomial Probabilities

Question:  `q001.  Note that there are 7 questions in this assignment.

What is the probability that on two rolls of a fair die, we obtain exactly two 3's?

Your solution:

Confidence Rating:

Given Solution:  The probability of obtaining a 3 on a single role is 1/6 (one of the six possible outcomes is a 3).  Since the two rolls are independent, it follows that if two dice are rolled the probability of obtaining two 3's is 1/6 * 1/6 = 1/36.

Self-critique:

Self-critique rating:

Question:  `q002.  What is the probability that on three rolls of a fair die, we obtain exactly two 5's?

Your solution:

Confidence Rating:

Given Solution:  On three rolls of a fair die, the two 5's can occur on the first and second, the first and third or the second and third rolls.  That is, of the three available positions the two positions in which the 5's occur can occur in C(3,2) = 3 different ways.

Since the probability of a 5 on any roll is 1/6 and the probability of not getting a 5 on a roll is 5/6. Any one of the three ways of getting two 5's and one non-5 is therefore (1/6) * (1/6) * (5/6 ) = 5/216.

Since each of the three ways to get the desired outcome occurs with probability 5/216, it follows that

Probability of exactly two 3's on three rolls = 3 * 5/216 = 15/216 = 5/72.

I’m confused as to why you multiply the 5/216 by 3? Doesn’t the 5/216 give you the correct answer? 1/6 *1/6 *5/6
shows the probability of all three dice.

INSTRUCTOR RESPONSE:

1/6 *1/6 *5/6 is the probability of getting 5 on the first die, 5 on the second and something else on the third.

However you can also get 5 on the first, something else on the second, and 5 on the third. The probability of this outcome could be written 1/6 * 5/6 * 1/6, showing the order of the three events.

Or you could get something else on the first and 5 on each of the last two. The probability of this outcome could be written 5/6 * 1/6 * 1/6, again showing the order of the three events.

When multiplied out, the probability of any of these three events is 1/6 * 1/6 * 5/6. The three events are mutually exclusive, so the probability that one of the three events will occur is 3 * 1/6 * 1/6 * 5/6.

Self-critique:

Self-critique rating:

Question:  `q003.  What is the probability that on six rolls of a fair die, we obtain exactly two 5's?

Your solution:

Confidence Rating:

Given Solution:  In order to get exactly two 5's on six rolls of the fair die, we must get two 5's and four results that are not 5.

The probability of getting a 5 on any roll is 1/6, and the probability of getting a result other than 5 is 5/6.

Therefore given any two positions out of the six the probability of obtaining 5's in two of the positions and non-5's in the remaining four positions is by the Fundamental Counting Principle

Probability of 5's in exactly two of the six positions = (1/6) * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) = (1/6)^2 * ( 5/6)^4.

There are C(6,2) ways in which the positions of the two 5's can be selected from the six available positions.  Thus we have

Probability of exactly two 5's on six flips = C(6,2) * (1/6)^2 * (5/6)^4.

STUDENT COMMENT:

So, because our outcome (two 5’s) can only occur 2 of the 6 times, in any order, we have to also multiply by C(6,2).
This is the same for the previous question, which was multiplied by 3 (or C(3,2) ) This concept makes a lot more sense
now.

INSTRUCTOR RESPONSE: 

Right. The two 5's can occur in any two of the 6 rolls, and there are C(6, 2) ways of selecting which two.

STUDENT COMMENT:  If I am reading this response correctly, I seem to have the correct idea in solving the problem above?

INSTRUCTOR RESPONSE: 

That is the right idea, but (1/6)^2 * (5/6)^4 would be the probability that you get the two 5's on two specified rolls
(e.g., the probability that you get the 5's on rolls 2 and 4).

You would also have a probability of (1/6)^2 * (5/6)^4 of getting the 5's on rolls 1 and 4, as well as the probability of
getting the 5's on rolls 2 and 5, etc..

You could get the 5's on rolls 1 and 2, or on rolls 1 and 3, or on rolls 1 and 4, or on rolls 1 and 5, or on rolls 1 and 6, or on rolls 2 and 3, or on rolls 2 and 4, or on rolls 2 and 5, or on rolls 2 and 6, or on rolls 3 and 4, or on rolls 3 and 5, or on rolls 3 and 6, or on rolls 4 and 5, or on rolls 4 and 6, or on rolls 5 and 6. That is C(6, 2) = 15 ways to get the two 5's, each with probability (1/6)^2 * (5/6)^4.

So the probability is

C(6, 2) * (1/6)^2 * (5/6)^5.

Self-critique:

Self-critique rating:

Question:  `q004.  If we let p stand for the probability of getting a 5 on a roll of a die and q for the probability of not getting a 5 on a roll, then how would we expressed a probability of getting exactly r 5's on n rolls?

Your solution:

Confidence Rating:

Given Solution:  By analogy with the preceding problem, we see that to get r 5's on n rolls we must get 5 the total of r times and non-5 a total of (n-r) times. 

Since probability of getting a 5 is p, the probability of getting 5 a total of r times is represented by p^r.

Since the probability of getting a non-5 is q, then the probability of getting a non-5 a total of (n-r) times is represented by q^(n-r).

There are C(n, r) ways to place fives in r of n positions, so the probability of getting 5 fives and n non-fives is

C(n, r) * p^r * q^(n-r).

STUDENT COMMENT:

I understood this problem pretty well but was still slightly confused about the q^(n-r) part.

INSTRUCTOR RESPONSE

On the previous example, n was 6 (for the 6 rolls) and r was 2 (the number of 'successful' outcomes, regarding a 5 as a success). The probability of a success was 1/6.

Our expression was C(6, 5) * (1/6)^2 * (5/6)^4. Using the symbols n, r and p we would write this as

• C(n, r) * p^r * (5/6)^4.

What about the 5/6 and the 4?

• 4 is the number of 'failures'. There were 6 rolls with 2 'successes' and therefore 6 - 2 = 4 'failures'.
  It should be clear that if there are n trials and r 'successes' there are (n - r) 'failures'.

• 5/6 in this example is the probability of 'not getting a 5', i.e., the probability of a 'failure'.
  The probability of a success and the probability of a failure add up to 1 (there is a probability of 1, or 100%, that the trial will either succeed or fail). If we let q stand for the probability of a failure, we get p + q = 1, so that q = 1 - p.

So our expression

C(n, r) * p^r * (5/6)^4 generalizes to

• C(n, r) * p^r * q^(n-r),

where q = 1 - p.

Self-critique:

Self-critique rating:

Question:  `q005.  Explain why, if p is the probability of getting a 5 on a single roll of a die, it follows that the probability q of not getting a 5 is  q = 1-p.

How would we therefore express the formula C(n, r) * p^r * q^(n-r) only in terms of p?

Your solution:

Confidence Rating:

Given Solution:  If we roll a single die, we either get 5 or we don't.  The two events are mutually exclusive -- they cannot both happen on the same roll.  They also cover all possibilities.  The sum of the probabilities is therefore 1. 

So we conclude that p + q = 1, and from this it follows immediately that q = 1 - p.

Substituting 1 - p for q in the expression C(n, r) * p^r * q^(n-r) we obtain

Probability of r fives on n rolls = C(n, r) * p^r * (1-p) ^ (n-r).

STUDENT COMMENT:

I have this wrong in the problem and am a bit confused. Why is the problem *p^r*(1-p)….?

INSTRUCTOR RESPONSE:

Recall that q = 1 - p. So

C(n, r) * p^r * q^(n-r) could be written as

C(n, r) * p^r * (1-p)^(n-r)

Self-critique:

Self-critique rating:

Question: `q006.  On five rolls of a die we could obtain two 3's on the first and second, or the first and third, or the first and fourth, or the first and fifth rolls. 

• List the other ways in which we could obtain two 3's.

• How many ways are there to obtain exactly two 3's?

• How would you express this result in terms of combinations, without the need to list all the ways?

What is each of the following probabilities?

• On the first roll you get a 3.

• On the second roll you get a 3.

• On the third roll you don't get a 3.

• On the fourth roll you don't get a 3.

• On the fifth roll you don't get a 3.

What therefore is the probability that you get 3's on each of the first two rolls, but not on any of the others?

What would be the probability that you get 3's on the first and third rolls, but not on any of the others?

We have listed all the ways you can get 3's on exactly two of the rolls.  Each of these ways has the same probability.  What therefore is the probability that, in one way or another, we get exactly two 3's?

Your solution:

Confidence Rating:

Self-critique Rating:

Question: `q007.  Explain why there are 15 different ways to get exactly two 5's on six rolls of a die.

Explain why the probability of each of these ways is 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6.

1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6 = 625 / 46656 (approximately .0134, or 1.34%). 

What therefore is the probability of getting exactly two 5's on six rolls of a die?

Your solution:

Confidence Rating:

Self-critique Rating: