Solution to Sample Test: MA 3210, Midterm 1, Spring 2000

Please note that I have dropped one more question on Adam's guilty questions.
Note 2: Do not do 13, 14, 15,16. Instead of 14, I expand the question 9 little more and moved to the end.
If you know how to do these, (this means that you are able to do the similar questions not the sample questions only), you will get least B. You also try to do as much as excercise problems as possible.

Good luck to you all!!

1.

(1) $A\cap B=\emptyset$. In other words, events A and B can not both occur simultanously.
(2) No. He needs to know $P(A \cap E)$.
(3) P(A|B) = P(A).
(4) In an experiment with a pair of dice, occurence of Die 1 and Die 2 is independent. Drawing a card with replacement is another example of independent events. The event that Long Island train is late today is independent to the event that Rolling Stone's concert in Paris is successful.
(5) $A \cup B \cup C = S$ and $A\cap B=\emptyset, A\cap C = \emptyset, B\cap C=\emptyset$.
(6) $P(A \cap E)=P(E) P(A\vert E)$.
(7) Yes. $P(A\cap B)= P(A) P(B)$.
(8) Yes. $P(A^{\prime} \vert E)= 1 - P(A\vert E)$.
(9) Yes. Since E,F froms a partion of S, P(A)=P(A|E) + P(A|F) so that

$$P(E\vert A) = \frac{E\cap A)}{P(A)} = \frac{P(E) P(A\vert)}{P(A\vert E) + P(A\vert F)}.$$

2.

Hint: Draw Venn Diagram for this sitation.
$P(M \cup H) = P(M) + P(H) - P(M \cap H) = 0.50 + 0.30 - 0.15 = 0.65$.
$P(M \cap H^{\prime}) = P(M) - P(M \cap M^{\prime}$.

3.

In the experiment of tossing a coin 5 times,
1. 2⁵=32.
2. $\left( \begin{array}{c} 5 \\ 2 \end{array} \right) = \frac{5!}{2! 3!} = 10$.
3. $\frac{ \left(\begin{array}{c} 5 \\ 2 \end{array} \right)}{32}=\frac{5}{16}$. You do not need to compute this expression.

4.

In a poker hand consisting of 5 cards, find the probability of holding
1.

$$\left( \begin{array}{c} 4 \\ 4 \end{array} \right) \left( \begin{array}{c} 48 \\ 1 \end{array} \right)\\ = 48$$

3 spades and 2 club:

$$\left( \begin{array}{c} 13 \\ 3 \end{array} \right) \left( \begin{array}{c} 13 \\ 2 \end{array} \right)$$

5.

Partitioning :

$$\left( \begin{array}{c} 6 \\ 1,~3,~2 \end{array} \right)= \frac{6!}{1! 3! 2!}=60$$

6.

Permutition with identical objects. 7 letters, two i, and two t.

$$\frac{7!}{2! 2!}$$

7.

P(H)=3/4, P(T) =1/4

. There are 2 outcomes which fit the description, say HT, TH. Each outcome has the probability of

$$\frac {3}{4}\cdot \frac {1}{4}= \frac{3}{16}. \mbox{ Hence the answer is} 2 \cdot \frac{3}{16} = \frac{3}{8}$$

8.

With the above biased coin:
(1) Did in class. Also See Figure 2.8 in the textbook; Replace Bag 1 with Coin tossing, the upper Bag 2 with 2 R and 2 B and the lower Bag 2 with 3 B, 1 R.
(2)

$$P(R\vert H)=\frac{2}{4}=\frac{1}{2}, P(R \cap H)= P(H) P(R\vert H) = \frac{3}{4} \frac{1}{2} = \frac{3}{8}.$$

For P(R), we use the fact that H, T form a partion. Hence by Theorem 2.16 , ie

$$P(R) = P(R\cap H) + P(R \cap T) = P(H) P(R \vert H) + P(T) P(R \cap T)$$

$$= 3/4 \cdot 2/4 + 1/4 \cdot 1/4 = \frac{7}{16}$$

(3) Bayes'Rule : But we already have P(R) from (2) so we have easy way;

$$P(H\vert R) = \frac{P(H \cap R) }{P(R)} = \frac{3/8}{7/16} = \frac{6}{7}$$

9.

It is moved to the end.

10.

We have

P(F₁) = 0.2, P(F₂) = 0.45, P(F₃) = 0.45

and

P(D|F₁) = 0.03, P(D|F₂) = 0.02, P(D|F₃) = 0.04

. (1) Use Theorem 2.16 with 3 factories which form a partion.

$$P(D) = P(D \cap F_1) + P(D \cap F_2) + P(D \cap F_3)$$

= P(F₁)P(D|F₁) + P(F₂) P(D|F₂) + P(F₃) P(D|F₃)

$$= 0.2 \cdot 0.03 + 0.3 \cdot 0.02 + 0.45 \cdot 0.04 = 0.006+ 0.006 + 0.018=0.03$$

(2)

$$P(F_1 \vert D) =\frac{F_1 \cap D) }{P(D)}= \frac{0.006 }{0.03} =0.2$$

11.

With the notation as in 12.,
G: A person is guilty
I: A person is innocent of the crime.
JI: Jury found the person innocent
JG: Jury found the person guilty.

P(JG|G)=0.8, P(JI|G) = 0.2, P(JG|I) = 0.01, P(G)=0.05

are given facts to us in order. I is a complement of G so that we have P(I) = 1- 0.05 =0.95

The probability we need is P(I | JG).

$$P(I \vert JG) = \frac{P(I) P(JG \vert I)}{P(G)P(JG\vert G) + ... ...rt I)} = \frac{0.95 \cdot 0.01 }{(0.05) (0.80) + (0.95) (0.01)}$$

12. We dropped this problem of Guilty data by Adam. This solution isfor

Formerly number 9: Two fair dice are tossed. Adam is interested in finding the probability of sum of the two dice and Judy is interested in finding the largest number of the two dice. (1) Find the probability that the sum is 10 for Adam.
(2) Find the probability that the sum is 8.
(3) Find the probability that the max number is 5 for Judy. (4) Adam and Judy want to use the term random variable. He decided to use X for his random variable and she decided to denote her random variable as Z. Describe X. (5) Find the outcomes (event) corresponding X=8 and find P(X=8)
(6) Find event whose random variable Z takes the value 4.

12. Question on Two fair dice are tossed. Part 1:
(1)3/36
(2) 5/36
(3) 9/36

Part 2:

(1) You can see from the table (4) below, but we will list them. (2,6), (3,5),((4,4)4),(5,3),(6,2) takes X=8. There are 5 of them and equally likely oucomes. So P(X=8) = 5/36
(2)Outcomes to X=3 are (1,2),(2,2) and (3,3) so that P(X=3)=3/36
(3)Outcomes to Z=4 are (1,4),(2,4),(3,4),(4,4),(4,3),(4,2),(4,1) of 7 outcomes and P(Z=4)=7/36.
(4), (5)

	X	Z		X	Z
(1,1)	2	1	(4,1)	5	4
(1,2)	3	2	(4,2)	6	4
(1,3)	4		(4,3)	7	4
(1,4)	5		(4,4)	8	4
(1,5)	6	5	(4,5)	9	5
(1,6)	7	6	(4,6)	10	6
(2,1)	3	2	(5,1)	6	5
(2,2)	4	2	(5,2)	7	5
(2,3)	5	3	(5,3)	8	5
(2,4)	6	4	(5,4)	9	5
(2,5)	7	5	(5,5)	10	5
(2,6)	8	6	(5,6)	11	6
(3,1)	4	3	(6,1)	7	6
(3,2)	5	3	(6,2)	8	6
(3,3)	6	3	(6,3)	9	6
(3,4)	7	4	(6,4)	10	6
(3,5)	8	5	(6,5)	11	6
(3,6)	9	6	(6,6)	12	6

x  or   z	f(x)	g(z)
1	0	1/36
2	1/36	3/36
3	2/36	5/36
4	3/36	7/36
5	4/36	9/36
6	5/36	11/36
7	6/36	0
8	5/36	0
9	4/36	0
10	3/36	0
11	2/36	0
12	1/36	0