# Probability HW Help

• May 14th 2006, 04:42 PM
Nimmy
Probability HW Help
1. A class is given a list of 20 study problems from which ten will be part of an upcoming exam. If a givent student knows how to solve 15 of the problems, find the probability that the student will be able to answer at least nine question on the exam.

2. Four letters and envelopes are addressed to four different people. If the letters are randomly inserted into the envelopes, what is the probability that (a)exactly one is inserted in the correct envelope and (b) at least one is inserted in the correct envelope?

3. A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random. What is the probablity that a) all four units are good? b) exactly two units are good? c) at least two units are good?

4. A sales representatives make sales at approximately one-fifth of all calls. If, on a given day, the representative contacts six potential clients, what is the probability that a sale will be made with (a) all six contacts, (b) none of the contacts, and (c) at least one contact?
• May 14th 2006, 07:19 PM
ThePerfectHacker
Quote:

Originally Posted by Nimmy
1. A class is given a list of 20 study problems from which ten will be part of an upcoming exam. If a givent student knows how to solve 15 of the problems, find the probability that the student will be able to answer at least nine question on the exam.

That means the probability that a student be able to answer any given question is $p=\frac{15}{20}=\frac{3}{4}$.
at least 9 mean probability of exactly 9 plus exactly 10 (all).
We have, for nine and ten respectively,
${10 \choose 9}(.75)^9(.15)=.1877$
${10 \choose 10}(.75)^{10}(.15)^0=.0563$
Thus,
$.0563+.1877=.2400$
• May 15th 2006, 10:55 AM
ThePerfectHacker
Quote:

Originally Posted by Nimmy

2. Four letters and envelopes are addressed to four different people. If the letters are randomly inserted into the envelopes, what is the probability that (a)exactly one is inserted in the correct envelope and (b) at least one is inserted in the correct envelope?

I did not find any elegant way of solving this problem.

I will solve an equivalent problem to this one. Consider a quadruple (A,B,C,D) where A,B,C,D are distinct digits of 1,2,3,4. We say one of these digits is 'correct' if it is its proper place from the left. For example, (1,2,3,4) all are correct (1,3,2,4) only two are correct because 1 and 4 are in thier proper places.
Hence, for problem (a) you need to find the probability that exactly once digits of A,B,C,D is correct. We do by using the fact that probability is the ratio of favorable outcomes to possible outcomes. Since the possible outcomes is 24 because for A we have 4 different numbers for B we have 3 because we used one for A and so on. Thus, we have $24=4\cdot 3\cdot 2\cdot 1$. Now we need to find the favorable outcomes. We do this in parts. We divide this problem into 4 parts. In the first part consider when A=1 (correct) and all the rest are wrong.
This is what you have (1,B,C,D) note that B can be either 3 or 4; C can be either 2 or 4; D can be either 2 or 3. If B=3 then, D=2 then C=4. If B=4 then C=2 then D=3. Thus there are only two possibilities when only the first one is correct. Now, do the same thing for (A,2,B,D) you will also get 2 possibilities and the same goes for (A,B,3,D) and (A,B,C,4) all give 2. In total you have 8 possible arrangents for exactly one of being correct. Thus, the probability is 8/24=1/3

(b)Note that,
$P(0)+P(1)+P(2)+P(3)+P(4)=1$
where $P(n)$ denotes the probability of exactly $n$. Note that at least 1 means,
$P(1)+P(2)+P(3)+P(4)$.
Thus, solving this equation we get,
$P(1)+P(2)+P(3)+P(4)=1-P(0)$.
Hence, find the probability of getting all wrong. Meaning that (A,B,C,D) are all wrong. Note that
A=2 or 3 or 4
B=1 or 3 or 4
C=1 or 2 or 4
D=1 or 2 or 3
If A=2 then D can take on two different values, which would force B and C to take on specific values.
If A=3 the D can take on two different values, which would force B and C to take on specific values.
If A=4 then B can take on two different values, which would force C and D to take on specific values.
In total you have 6 arrangents. Thus, the probability is 6/24=1/4.

Hence,
$P(1)+P(2)+P(3)+P(4)=1-1/4=\frac{3}{4}$
• May 15th 2006, 11:02 AM
ThePerfectHacker
Quote:

Originally Posted by Nimmy
3. A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random. What is the probablity that a) all four units are good? b) exactly two units are good? c) at least two units are good?

This is a straight out binomial probability problem. Taking any one the probability is 3/12=1/4 and 4 are taken.

(a) ${4 \choose 4}(.75)^4(.25)^0=.3164$
(b) ${4 \choose 2}(.75)^2(.25)^2=.2109$
(c) $\sum_{k=2}^4{4 \choose k}(.75)^k(.25)^{n-k}=.9492$
• May 15th 2006, 11:08 AM
ThePerfectHacker
Quote:

Originally Posted by Nimmy

4. A sales representatives make sales at approximately one-fifth of all calls. If, on a given day, the representative contacts six potential clients, what is the probability that a sale will be made with (a) all six contacts, (b) none of the contacts, and (c) at least one contact?

Just problem 3 only here the probability is 1/5.
Thus,
(a) ${6 \choose 6}(.20)^6(.80)^0=0.000064$

(b) ${6 \choose 0}(.20)^0(.80)^6=.2621$

For (c) do the NOT, meaning, find the probability of selecting none which we got in the previous problem as .2621. Thus, the probability is 1-.2621=.7379