Lesson Challenging probability problems

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<H2>Challenging probability problems</H2>

<H3>Problem 1</H3>Each of &nbsp;9&nbsp; different books is to be given at random to &nbsp;Rebecca, &nbsp;Sheila, &nbsp;or &nbsp;Tom. 
What is the probability that &nbsp;Rebecca receives &nbsp;2&nbsp; books, &nbsp;Sheila receives &nbsp;4, &nbsp;and &nbsp;Tom receives &nbsp;3?

<B>Solution</B>

<pre>
1.  First of all, let me note/clarify that the condition does not prohibit that all 9 books go to Rebecca.

It also  does not prohibit that all 9 books go to Sheila.

It also does not prohibit that all 9 books go to Tom.

Then the total number of different ways to distribute 9 books between 3 persons is {{{3^9}}}.

To better understand why it is so, imagine, that the teacher observes this process and write to his notebook to whom of the three persons

the current book goes. He write "R" if the book goes to Rebecca; writes "S" if the book goes to Sheila; and write "T" if the book goes to Tom.

So, for each way of distribution of 9 books the teacher has written the word of the length 9, consisting of 3 letters R, S and/or T.

Repetitions are allowed - so the number of such words is {{{3^9}}}, as I said above.

2.  Now we can select 2 books for Rebecca from 9 books by {{{C[9]^2}}} = {{{(9*8)/2}}} = 36 ways.

We can select 4 books for Sheila from remaining (9-2) = 7 books by {{{C[7]^4}}} = {{{(7*6*5))/(1*2*3)}}} = 35 ways.

As soon as the books for Rebecca and for Sheila are selected, Tom will take the remaining books - he has no choice.

3.  Therefore, the probability under the question is equal to  {{{(36*35)/3^9}}} = 0.064 = 6.4%.
</pre>

<H3>Problem 2</H3>What is the probability that a randomly chosen four-digit number contains no repeated digits?

<B>Solution</B>

<pre>
1.  The full space of events consists of 9999-999 = 9000 elements (4-digit numbers).

2.  Of them, the favorable sub-space are those 4-digit numbers that have no repeating digits.

The number of elements in this sub-space = 9*9*8*7

(any of 9 digits from 1 to 9 in the left-most position;   (0 (zero) is excluded);

any of remaining 9 digits in the next position;   (0 is included !)

. . . and so on . . .

3.  Hence, the probability under the question is  {{{(9*9*8*7)/9000}}} = 0.504 = 50.4%.
</pre>

<H3>Problem 3</H3>What is the probability to select randomly a code which doesn't contains &nbsp;7&nbsp; among all possible &nbsp;4&nbsp; digit identification codes?

<B>Solution</B>

<pre>
Since otherwise is not stated, I will assume that all 10 digits from 0 to 9 are allowed for all positions of identification codes

and repeating of digits is allowed.

Then the total number of all possible codes is equal to  {{{10^4}}}  (each of ten digits is allowed in each of 4 positions).

The number of codes that do not the digit of 7 in any of 4 positions is  {{{9^4}}}  (9 digits in each of 4 positions).

Therefore the probability under the question is  {{{9^4/10^4}}} = {{{(9/10)^4}}} = 0.6561 = 65.61%.
</pre>

<H3>Problem 4</H3>A bag contains &nbsp;7&nbsp; red marbles, &nbsp;9&nbsp; white marbles, &nbsp;and &nbsp;9&nbsp; blue marbles. 
You draw &nbsp;5&nbsp; marbles out at random, &nbsp;without replacement.

What is the probability that all the marbles are red? 
What is the probability that exactly two of the marbles are red? 
What is the probability that none of the marbles are red?

<B>Solution</B>

Notice that the total number of marbles is &nbsp;7 + 9 + 9 = 25.

Therefore the expressions for probabilities in all &nbsp;3&nbsp; cases have  &nbsp;&nbsp;{{{C[25]^5}}}  &nbsp;&nbsp;as the denominator.

<pre>
(a)  What is the probability that all the marbles are red?

You can select 5 red marbles of 7 red marbles in  {{{C[7]^5}}} ways. Therefore the answer is  {{{C[7]^5/C[25]^5}}}.

(b)  What is the probability that exactly two of the marbles are red?

You can select 2 red marbles of 7 red marbles in  {{{C[7]^2}}} ways.  Then you can add any 3 marbles of remaining 9+9 = 18 white and blue marbles.

You can do it in {{{C[7]^2*C[18]^3}}} ways.  Therefore the answer is  {{{(C[7]^2*C[18]^3)/C[25]^5}}}  ways.

It means that you select 5 marbles among 18 = 9 + 9 white and blue marbles.

You can do it in  {{{C[18]^5}}} ways.  Therefore, the answer is  {{{C[18]^5/C[25]^5}}}.  
</pre>

<H3>Problem 5</H3>On a table in a doctor's office are five Motor Trend magazines, &nbsp;six Parenting magazines, &nbsp;and three Sports Illustrated magazines. 
If a patient randomly selects three magazines, &nbsp;determine the probability that

a. &nbsp;three Motor Trend magazines are selected.
b. &nbsp;two Parenting magazines and one Sports Illustrated magazine are selected.
c. &nbsp;no Parenting magazine is selected.
at &nbsp;least one Parenting magazine is selected.

<pre>
First notice that the total number of magazines = 5 + 6 + 3 = 14.

a. three Motor Trend magazines are selected.    Probability = {{{C[5]^3/C[14]^3}}}.

b. two Parenting magazines and one Sports Illustrated magazine are selected.   Probability = {{{(C[6]^2*C[3]^1)/C[14]^3}}}.

c. no Parenting magazine is selected.   Probability = {{{C[8]^3/C[14]^3}}}.

Here 8 = 14-6 is the number of magazines on the table excluding 6 Parenting magazines.

d. at least one Parenting magazine is selected.   Probability = complement to the "c"Probability.
</pre>

<H3>Problem 6</H3>A bag contains &nbsp;10&nbsp; red, &nbsp;8 &nbsp;green and &nbsp;7 &nbsp;blue balls. &nbsp;If two balls are picked at random one after the other without replacement from the bag, 
what is the probability that: 
(i) &nbsp;&nbsp;they are of the different colors ?
(ii) &nbsp;&nbsp;at least one ball is blue ?

<B>Solution</B>

(i) &nbsp;&nbsp;they are of the different colors ?

<pre>

The favorable sets are  (R,G) or (R,B) or (G,B).

Correspondingly, the probability under the question is

P = {{{(10/25)*(8/24)}}} + {{{(10/25)*(7/24)}}} + {{{(8/25)*(7/24)}}}.

Here 25 = 10 + 8 + 7 is the total number of balls and 24 = 25-1.

You may complete it as a fraction or use your calculator to get the decimal number.
</pre>

(ii) &nbsp;&nbsp;at least one ball is blue?

<pre>
     The favorable sets are  (R,B) or (G,B) or (B,B).

Correspondingly, the probability under the question is

P = {{{(10/25)*(7/24)}}} + {{{(8/25)*(7/24)}}} + {{{(7/25)*(6/24)}}}.

You may complete it as a fraction or use your calculator to get the decimal number.
</pre>

<H3>Problem 7</H3>Keith's Florists has &nbsp;15&nbsp; delivery trucks, &nbsp;used mainly to deliver flowers and flower arrangements 
in the Greenville, &nbsp;South Carolina, &nbsp;area. &nbsp;Of these &nbsp;15&nbsp; trucks, &nbsp;7&nbsp; have brake problems. 
A sample of &nbsp;6&nbsp; trucks is randomly selected. &nbsp;What is the probability that &nbsp;2&nbsp; of those tested have defective brakes?

<B>Solution</B>

<pre>
This probability is this ratio  {{{(C[7]^2*C[8]^4)/C[15]^6}}}.

We start choosing 2 defective cars among of 7 defective. It can be done in  {{{C[7]^2}}} = {{{(7*6)/2}}} = 21 ways.

We complement 2 defective cars to 6 by adding 4 regular cars from 8 = 15-7 regular cars.  It can be done by  {{{C[8]^4}}} = {{{(8*7*6*5)/(1*2*3*4)}}} = 70 ways.

The total number of ways to choose 6 cars of 15 is  {{{C[15]^6}}} = {{{(15*14*13*12*11*10)/(1*2*3*4*5*6)}}} = 5005.

Thus the probability under the question is  {{{(21*70)/5005}}} = 0.2937 = 29.37%.
</pre>

<H3>Problem 8</H3>A quick quiz consists of 3 multiple choice problems, &nbsp;each of which has &nbsp;5&nbsp; answers, &nbsp;only one of which is correct. 
If you make random guesses on all &nbsp;3&nbsp; problems,

What is the probability that all &nbsp;3&nbsp; of your answers are incorrect ?
What is the probability that all &nbsp;3&nbsp; of your answers are correct ?

<B>Solution</B>

<pre>
    Imagine that for each of the 3 multiple choice question the answers are labeled by 5 letters A, B, C, D and E

(5 possible answers to each question).

By answering to each question, you mark your answer by one of the 5 letters.

So, by answering to 3 questions, you write the word of the length 3, using one of 5 letters in each of the three positions.

It is your model. The entire space of events consists of all {{{5^3}}} = 125 such words of the length 3, written in 5-letter alphabet.

Now we are ready to answer the problem's questions.
</pre>

(a)  &nbsp;&nbsp;<U>What is the probability that all &nbsp;3&nbsp; of your answers are incorrect ?</U>

<pre>
     If all 3 of your answers are incorrect, it means that in each of the 3 positions you put one of 4 letters, distinct of correct.

You can do it in  {{{4^3}}} ways, therefore, the probability under the question is  {{{4^3/5^3}}} = {{{(4/5)^3}}} = {{{0.8^3}}} = 0.512.
</pre>

(b)  &nbsp;&nbsp;<U>What is the probability that all &nbsp;3&nbsp; of your answers are correct ?</U>

<pre>
     If all 3 of your answers are correct, it means that in each of the 3 positions you guessed the correct letter.

There is <U>ONLY ONE way</U> to do it; therefore, the probability under the question is  {{{1/5^3}}} = {{{1/125}}} = 0.008.
</pre>

<H3>Problem 9</H3>A single die is rolled &nbsp;22 &nbsp;times.

What is the probability that a six is rolled exactly once, &nbsp;if it is known that at least one six is rolled?

<B>Solution</B>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Let us solve the problem step by step.

<pre>
1.  The probability that a six is rolled exactly once, if it is known that at least one six is rolled, is equal to the ratio,

where the numerator is the probability that a six is rolled exactly once,

while the denominator is the probability that at least one six is rolled.

2.  The numerator is equal to  {{{C[22]^1*(1/6)*(5/6)^(22-1)}}} = {{{22*(1/6)*(5/6)^21}}}.

3.  The denominator is the complement to the probability that no one six is rolled.

The probability that no one six is rolled is  {{{(5/6)^22}}};  hence, the complement to it is  1 - {{{(5/6)^22}}}.

4.  Therefore, the probability under the question is

{{{(22*(1/6)*(5/6)^21)/(1 - (5/6)^22)}}} = 0.081172 = 8.1172%  (approximately).     <U>ANSWER</U>
</pre>

<H3>Problem 10</H3>A bird breeder has a total of &nbsp;50 &nbsp;canaries, &nbsp;21 &nbsp;of which are
male and &nbsp;29 &nbsp;of which are female. &nbsp;A pet shop randomly
selects 8 of them.

<pre>
    1. find the probability that none of the 8 are female.
    2. find the probability that exactly 5 of the 8 are male.
    3. find the probability that at least 6 of the 8 are female.
    4. find the probability that at least 2 of the 8 are male.
</pre>

<B>Solution</B>

<pre>
1.  It is the same as the probability that all 8 are males.

P = {{{C[21]^8/C[50]^8}}},  or, which is the same,

P = {{{(21/50)*(20/49)*(19/48)*(18/47)*(17/46)*(16/45)*(15/44)*(14/43)}}}.

They are not only "different formulas" - they go with different explanations (=self-explanations),

but they produce identical results.

2.  P = {{{(C[21]^5*C[29]^3)/C[50]^8}}}.

3.  P (at least 6 of the 8 are females) = P(0 males) + P(1 male) + P(2 males) = {{{C[29]^8/C[50]^8}}} + {{{21*C[28]^7/C[50]^8}}} + {{{C[21]^2*C[27]^6/C[50]^8}}}.

4.  P (at least 2 of the 8 are males)   = complement to P(at most 1 of the 8 is male) = 1 - P(0 males) + P(1 male) = 1 - {{{C[29]^8/C[50]^8}}} - {{{(C[28]^7*21)/C[50]^8}}}.
</pre>

My other lessons on &nbsp;<B>Probability</B>&nbsp; in this site are

<TABLE>
  <TR>
  <TD>&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Simple-and-simplest-probability-problems.lesson>Simple and simplest probability problems</A> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Solving-probability-problems-using-complementary-probability.lesson>Solving probability problems using complementary probability</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Elementary-Probability-problems-related-to-Combinations.lesson>Elementary Probability problems related to combinations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/A-True-False-test.lesson>A True/False test</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/A-multiple-choice-answers-test.lesson>A multiple choice answers test</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Coinciding-birthdays.lesson>Coinciding birthdays</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/A-shipment-containing-fair-and-defective-alarm-clocks-.lesson>A shipment containing fair and defective alarm clocks</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/People-in-a-room-write-down-integer-numbers-at-random.lesson>People in a room write down integer numbers at random</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/A-drawer-contains-a-mixture-of-socks.lesson>A drawer contains a mixture of socks</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Students-studying-foreign-languages.lesson>Students studying foreign languages</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-for-a-computer-to-be-damaged-by-viruses.lesson>Probability for a computer to be damaged by viruses</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Conditional-probability-problems.lesson>Conditional probability problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Using-sample-space-to-solve-Probability-problems.lesson>Using sample space to solve Probability problems</A>
 </TD>
  <TD>
&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Typical-probability-problems-from-the-archive.lesson>Typical probability problems from the archive</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Geometric-probability-problems.lesson>Geometric probability problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Advanced-probability-problems-from-the-archive.lesson>Advanced probability problems from the archive</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Selected-probability-problems-from-the-archive.lesson>Selected probability problems from the archive</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Experimental-probability-problems.lesson>Experimental probability problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Rolling-a-pair-of-fair-dice.lesson>Rolling a pair of fair dice</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Ten-pairs-of-shoes-are-kept-in-a-rack.lesson>Ten pairs of shoes are kept in a rack</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/A-diplomat-and-spies.lesson>A diplomat and spies</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Elementary-operations-on-sets-help-solving-Probability-problems.lesson>Elementary operations on sets help solving Probability problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Independent-and-mutually-exclusive-events.lesson>Independent and mutually exclusive events</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Unusual-probability-problems.lesson>Unusual probability problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-problem-for-the-Day-of-April-1.lesson>Probability problem for the Day of April, 1</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/OVERVIEW-of-lessons-on-Probability.lesson>OVERVIEW of lessons on Probability</A>

</TD>
 </TR>
</TABLE>

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-II.