stat hw

STAT 301.001 Fall 2016, Homework 5

Due: September 30th, Friday at the beginning of class

Show your work in the answers.

You may keep the form of ( n k

) or k! as the final answer unless stated

otherwise.

Problem 1

Find the values (in the form of integers) of the following expressions.

(a) 4! (b) (5)4 (c) (100)2 (d) ( 5 3

) (e)

( 100 98

)

Problem 2

A sequence consisting of only 0’s and 1’s is called a binary sequence, e.g. 0110111001.

(a) Find the number of different binary sequences of length 6.

(b) Find the number of different binary sequences of length 6 in which exactly three of the digits are 0.

(c) Find the number of different binary sequences of length 6 in which at least one of the digits is 0.

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Problem 3

In this problem, an example of using product principle incorrectly will be shown. Please read it carefully and keep it in mind when you use the product principle in the future.

A hand of 5 cards is drawn from 52 cards. Our goal in this problem is to find the probability of the event

A = {the five-card hand contains at least two aces and two 7’s}.

The sample space Ω is the set of all possible five-card hand, as shown in Page 7 in Lecture 13 slides.

The following is a WRONG way to use the product principle to count the number of outcomes in this event:

The task can be divided into 3 steps:

Step 1. Choose 2 cards of rank ace.

Step 2. Choose 2 cards of rank 7.

Step 3. Choose 1 card from the rest cards.

Then by the product principle, the number of outcomes is the product of the number of different choices in the three steps.

The above procedure is wrong, because it may count the same outcome multiple times. Choosing differently in the steps may lead to the same result. For example, the following two different combinations of choices

• Step 1: Choose ♠A and ♥A;

Step 2: Choose ♠7 and ♥7;

Step 3: Choose ♦A.

• Step 1: Choose ♠A and ♦A;

Step 2: Choose ♠7 and ♥7;

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Step 3: Choose ♥A.

result in the same hand ♠A♥A♦A♠7♥7. As a result, the answer from the wrong procedure may be larger than the correct number.

The main reason that different ways may lead to the same outcome is that the card chosen in Step 3 may be an ace or 7. Therefore, for the correct answer, we may first partition the event A into three events, depending on what the “5th card” is:

A1 = {the five-card hand contains three aces and two 7’s} A2 = {the five-card hand contains two aces and three 7’s} A3 = {the five-card hand contains exactly two aces and exactly two 7’s}

By addition principle, #A = #A1 + #A2 + #A3. By symmetry, #A1 and #A2 should be the same.

You are asked to complete the rest of the calculation:

(a) Find #Ω

(b) Find #A1.

(c) Find #A3.

(d) Find #A and P (A).

Problem 4

(continued from Homework 4)

Suppose that tomorrow will be either a rainy day or sunny day, and you may go hiking tomorrow. The probability that tomorrow will be a rainy day is 0.4. The probability that you will go hiking tomorrow is 0.8. The probability that you will go hiking tomorrow on a rainy day is 0.3.

(a) What is the probability that you will go hiking tomorrow given that it is a rainy day tomorrow?

(b) What is the probability that you will go hiking tomorrow given that it is a sunny day tomorrow?

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(c) What is the probability that tomorrow will be a rainy day given that you will not go hiking tomorrow?

(d) Is the two events “You will go hiking tomorrow” and “It is a rainy day tomorrow” independent?

Problem 5

A hand of 5 cards is drawn from 52 cards. Suppose that after you draw the first two cards, you see that they are ♠A and ♠7.

(a) The sample space Ω is the same as in Problem 3. Let

A = {The hand consists of ♠A and ♠7}

What is the number of different hands in A?

(b) What is the probability that the 5-card hand is a flush (all 5 cards are of the same suit) given that two cards are ♠A and ♠7 ?

Hint: Define B = {The hand is a flush}. Then

A∩B = {The hand is a flush consists of ♠A and ♠7}

The counting task is only to determine the other three ♠ cards.

(c) What is the probability that the 5-card hand is a full house (3 cards are of the same rank and the other two are of another rank) given that two cards are ♠A and ♠7 ?

Hint: For the intersection event, you may need to partition it into two disjoint events. Review Problem 3 if necessary. For the event of drawing a full house, the number of possible outcomes in this event is already calculated in class, so you can use the result directly in this problem.

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