# Probability Conundrums

#### MrGray

I've been practicing probability questions for a soon to arrive entrance exam and some of them seem to completely stump me.

1) In a horse race where 18 horses numbered 1-18. The probability that horse 1 would win is 1/6, that 2 would win is 1/10 and that 3 would win is 1/8. Assuming that a tie is impossible, find the change that one of the three will win.

a) 47/120
b) 119/120
c) 11/129
d) 1/5
e) None of these

2) A and B are two candidates seeking admission to an Ivy League college. The probability that A is selected is 0.5 and the probability that A and B are selected is at most 0.3. Is it possible that the probability of B getting selected is 0.9?

a) No
b) Yes
c) Either (a) or (b)
d) Can't say

3) The probability that a student will pass in Mathematics s 3/5 and the probability that he will pass in English is 1/3. If the probability that he will pass in both Mathematics and English is 1/8, what is the probability that he will pass in at least one subject?

a) 97/120
b) 87/120
c) 53/120
d) 120/297

4) The odds in favor of standing first of three students A, B and C appearing at an examination are 1:2, 2:5 and 1:7 respectively. What is the probability that either of them will stand first (assume that a tie for the first place is not possible.)

a) 168/178
b) 122/168
c) 5/168
d) 125/168

5) A and B are two mutually exclusive and exhaustive events associated with a random experiment. Find P(A) if it is given that P(B) = 3/2 x P(A) and P(C) = 1/2 x P(B).

a) 0.25
b) 0.3
c) 0.1
d) 0.2

6) If P(A) = 1/3, P(B) = 1/2, P(A intersection B) = 1/4 then find P(A' union B').

a) 1/3
b) 2/5
c) 2/3
d) 3/4

Any help would be greatly appreciated!

#### Plato

MHF Helper
We will not simply work them out!
So you have to tell us what you have done.
Where do you have trouble?

• mr fantastic

#### Will

Re question 3

Personally I don't find the fractional probabilites that helpful. Try to think of it like this. The answer has to be between 0 and 1 or whatever the fractional equivilent is.

There are 3 possibilities. Pass 1 exam $$\displaystyle E1$$ pass 2 exams $$\displaystyle E2$$ or fail both $$\displaystyle F2$$ If you add the probabilities of all those together they add up to 1. So to work out the probability of passing one exam subtract the probabilities of passing both and failing both from 1. What remains has to be the probability of passing 1 exam.

$$\displaystyle P(E1) = 1 - P(E2)+(F2)$$

With question 3 you have the probability that they will pass both exams and the probability of passing each individual exam. To work out the probability of failing both subtract the probability of passing each exam from 1 and then multiply the answers together.

*Note i'm a newbie so the probability that this is wrong has to be greater than 0 so I would appreciate help from other forum members to check if this is right.

#### Plato

MHF Helper
Re question 3
There are 3 possibilities. Pass 1 exam $$\displaystyle E1$$ pass 2 exams $$\displaystyle E2$$ or fail both $$\displaystyle F2$$ If you add the probabilities of all those together they add up to 1. So to work out the probability of passing one exam subtract the probabilities of passing both and failing both from 1. What remains has to be the probability of passing 1 exam.
$$\displaystyle P(E1) = 1 - P(E2)+(F2)$$
With question 3 you have the probability that they will pass both exams and the probability of passing each individual exam. To work out the probability of failing both subtract the probability of passing each exam from 1 and then multiply the answers together.
*Note i'm a newbie so the probability that this is wrong has to be greater than 0 so I would appreciate help from other forum members to check if this is right.
This is totally off the mark.

#### MrGray

We will not simply work them out!
So you have to tell us what you have done.
Where do you have trouble?
Apologies,

1) I tried to solve this by finding the probability in the following cases,

1 2 3
W L L
L W L
L L W

The manifest flaw is of course that I have not taken into consideration the other 15 horses - which I can't find a way to.

2) Never mind, got this.

3) P(M) = 3/5
P(E) = 1/3
So shouldn't P(M & E) = 1/5?
Couldn't figure out what 1/8 was and how it equated itself in the final answer.

4) Thus,
P(A) = 1/3
P(B) = 2/7
P(C) = 1/8

Calculated for the following cases,

A B C
F NF NF
NF F NF
NF NF F

5) P(A) = 2/3 x P(B)
P(B) = 3/2 x P(A)
P(C) = 1/2 x 3/2 x P(A)

That's about as far as I got.

6) Set theory and its correlation with probability is lost on me here.

Regards,
AG

#### Plato

MHF Helper
In #1 you are simply asked “what is the probability that horse 1, 2 or 3 will win?”
With no ties that is simply $$\displaystyle \dfrac{1}{6}+\dfrac{1}{10}+ \dfrac{1}{8}$$.

For #3 , find $$\displaystyle P(M\cup E)=P(M)+P(E)-P(M\cap E)$$.

#6. You have no business trying probability without knowing basic set theory.
$$\displaystyle P\left( {A' \cup B'} \right) = P\left( {A'} \right) + P\left( {B'} \right) - P\left( {A' \cap B'} \right)$$
$$\displaystyle = \left[ {1 - P(A)} \right] + \left[ {1 - P(B)} \right] - \left[ {1 - P(A \cup B)} \right]$$

• MrGray

#### MrGray

In #1 you are simply asked “what is the probability that horse 1, 2 or 3 will win?”
With no ties that is simply $$\displaystyle \dfrac{1}{6}+\dfrac{1}{10}+ \dfrac{1}{8}$$.

For #3 , find $$\displaystyle P(M\cup E)=P(M)+P(E)-P(M\cap E)$$.

#6. You have no business trying probability without knowing basic set theory.
$$\displaystyle P\left( {A' \cup B'} \right) = P\left( {A'} \right) + P\left( {B'} \right) - P\left( {A' \cap B'} \right)$$
$$\displaystyle = \left[ {1 - P(A)} \right] + \left[ {1 - P(B)} \right] - \left[ {1 - P(A \cup B)} \right]$$
#1 - should have seen that, thanks.

#3 and #6 - understood. Guess I need to oil set theory.

Anything on #4 and #6?

#### mr fantastic

MHF Hall of Fame
#1 - should have seen that, thanks.

#3 and #6 - understood. Guess I need to oil set theory.

Anything on #4 and #6?
#6: You said in a previous post that you understood #6. What's changed?

#4: Use an appropriate formula for $$\displaystyle \Pr(A \cup B \cup C)$$ - sure to be somewhere in your class notes or textbook. In fact, all of these questions rely on your ability to translate the words into basic probaility statements and then manipulate and apply basic formulae. You are well-advised to go back and thoroughly review all this material, including working through worked examples.

In future, please don't post more than two questions in a thread. Otherwise the thread can get convoluted and difficult to follow. It is better for forum organization and better for you to get your questions answered in a more timely manner if you start new threads as necessary for remaining questions. eg. If you have five questions, post two of them in two threads and start a new thread for the remaining one etc.

And if the question has more than two parts to it, it is best to post only that question and its parts in the thread and start a new thread for other questions.

#### MrGray

#6: You said in a previous post that you understood #6. What's changed?
Meant to write #5, but never mind - figured this one out.

ro#4: Use an apppriate formula for $$\displaystyle \Pr(A \cup B \cup C)$$ - sure to be somewhere in your class notes or textbook. In fact, all of these questions rely on your ability to translate the words into basic probaility statements and then manipulate and apply basic formulae. You are well-advised to go back and thoroughly review all this material, including working through worked examples.

In future, please don't post more than two questions in a thread. Otherwise the thread can get convoluted and difficult to follow. It is better for forum organization and better for you to get your questions answered in a more timely manner if you start new threads as necessary for remaining questions. eg. If you have five questions, post two of them in two threads and start a new thread for the remaining one etc.

And if the question has more than two parts to it, it is best to post only that question and its parts in the thread and start a new thread for other questions.
Aye and thanks for the help.