Lesson Probability problems similar to a coinciding birthdays problem
Algebra
->
Probability-and-statistics
-> Lesson Probability problems similar to a coinciding birthdays problem
Log On
Algebra: Probability and statistics
Section
Solvers
Solvers
Lessons
Lessons
Answers archive
Answers
Source code of 'Probability problems similar to a coinciding birthdays problem'
This Lesson (Probability problems similar to a coinciding birthdays problem)
was created by by
ikleyn(52780)
:
View Source
,
Show
About ikleyn
:
<H2>Probability problems similar to a coinciding birthdays problem</H2> <H3>Problem</H3>John was trying to bet Krista $100 that he could pick 7 people at random and at least two of them would have a birthday in the same month. Find the probability for each of them to win. <B>Solution</B> <pre> Consider 7 people A, B, C, D, E, F and G. Person A has birthday in some of 12 months, X. The probability that person B has birthday in some other month Y, different from X, is {{{11/12}}}. The probability that person C has birthday in some other month Z, different from X and Y, is {{{10/12}}}. The probability that person D has birthday in some other month W, different from X, Y and Z, is {{{9/12}}}. Thinking this way, we get that the probability for 7 persons to have birthdays in different months is {{{(11/12)*(10/12)*(9/12)*(8/12)*(7/12)*(6/12)}}}, which is the product of 6 factors; the structure of the formula is clear. Making calculations, you get value P = 0.111400463, or 0.11 = 11% rounded. So, the probability that 7 randomly chosen people have birthdays in different months is 0.11, or 11%. The probability that of 7 randomly chosen people at least two of them have birthdays in the same month is the complement to 0.11, i.e. 1-0.11 = 0.89. Thus the probability that John wins is 0.89, or 89%; the probability that Krista wins is 0.11, or 11%. <U>ANSWER</U>. The probability that John wins is 0.89 = 89%. The probability that Krista wins is 0.11 = 11%. </pre> <H3>Problem 2</H3>What is the probability that if 11 letters are typed, no letters are repeated? <B>Solution</B> <pre> We have 26 lowercase letters and 26 uppercase letters in English alphabet. Assuming that we use lowercase letters only, there are {{{26^11}}} possible words of the length 11. Of them, the number of words with no repeating letters is 26*25*24*23*22*21*20*19*18*17*16 (11 factors of integer numbers in descending order starting from 26, in all). The probability do not have repeating letters is 1 - {{{(25/26)*(24/26)*(23/26)*(22/26)*(21/26)*(20/26)*(19/26)*(18/26)*(17/26)*(16/26)}}} = = 1 - 0.084026 = 0.915974 (rounded). <U>ANSWER</U> </pre> <H3>Problem 3</H3>The elevator at the Matthews Building of the University of New South Wales starts with 10 passengers at the ground and makes stops at each of the 14 floors above. Assuming that it is equally likely that a passenger gets off at any of these 14 floors, what is the probability that at least two of these passengers will get off at the same floor? <B>Solution</B> <pre> For 10 passengers, make all possible lists/schedules, at which of 14 floors above they can get off. 1st passenger can get off at any of 14 floors above, giving 14 options. 2nd passenger can get off at any of 14 floors above, giving 14 options. 3rd passenger can get off at any of 14 floors above, giving 14 options. . . . . . . . and so on . . . . . . . 10th passenger can get off at any of 14 floors above, giving 14 options. So, in all, there are {{{14^10}}} different possible schedules. Next, let's calculate, how many of these schedules have different floors for different people. 1st passenger can get off at any of 14 floors above, giving 14 options. 2nd passenger can get off at any of remaining 13 floors, giving 13 options. 3rd passenger can get off at any of remaining 12 floors, giving 12 options. . . . . . . . and so on . . . . . . . 10th passenger can get off at any of remaining 14-9 = 5 floors, giving 5 options. So, the number of schedules that have have different floors for different people is 14*13*12*. . . *5. Thus, the probability that no two passengers get out on some floor is {{{(14/14)*(13/14)*(12/14)*ellipsis*(5/14)}}} = 0.01256 (rounded). From here, the probability that at least two of these passengers will get off at the same floor is the COMPLEMENT of this value to 1 P = 1 - {{{(14/14)*(13/14)*(12/14)*ellipsis*(5/14)}}} = 1 - 0.01256 = 0.98744 (rounded). <U>ANSWER</U>. The probability that at least two of these passengers will get off at the same floor is P = 1 - {{{(14/14)*(13/14)*(12/14)*ellipsis*(5/14)}}} = 1 - 0.01256 = 0.98744 (rounded) </pre> <H3>Problem 4</H3>There are 10 students with different designs of gloves. A professor asked them to put all their gloves in a box and he randomly selected 6 pieces of gloves. What is the probability that there is exactly one matching pair of gloves? <B>Solution</B> <pre> There are 20 separate pieces in the box (of 10 pairs of gloves). The probability the problem asking for, is the ratio, whose denominator is the number of all sextuples that can be selected from the set of 20 items, {{{C[20]^6}}} = 38760. The numerator is the number of all sextuples that have one matching pair and 4 non-matching items. Our task is to calculate the number of such sextuples. In turn, this number is the product of 10 (the number of gloves) multiplied by the number of all possible quadruples that consist of non-matching pairs from remaining 2*(10-1) = 18 items. Thus our task is to calculate the number of all non-matching quadruples of 18 separate pieces. We can select 1st item as any of 18 pieces, which gives us 18 options. After that, we can choose next item from remaining 18-2 = 16 items (excluding item #1 and its twin). It gives us 16 possible options. After that, we can choose 3rd item from remaining 16-2 = 14 items (excluding item #1 and its twin, as well as item #2 and its twin). It gives us 14 possible options. After that, we can choose 4th item from remaining 14-2 = 12 items (excluding item #1 and its twin, item #2 and its twin, item #3 and its twin). It gives us 12 possible options. So, the number of all different quadruples of non-matching items from the total of 18 items (9 gloves) is 18*16*14*12 = 48384. Notice that 48384 is the number of all such different {{{highlight(ordered)}}} quadruples, so we need to divide this number by 4! = 24, to count {{{highlight(unordered)}}} quadruples. Thus, the number of unordered quadruples is 48384/24 = 2016, and the number of all sextuples we are interesting to count is 10*2016 = 20160. Now the probability we are looking for is P = {{{20160/38760}}} = {{{168/323}}} = 0.5201 (rounded). <U>ANSWER</U> </pre> My other lessons on <B>Probability</B> in this section are - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Probabilistic-analysis-of-a-court-verdicts.lesson>Probabilistic analysis of a court verdicts</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/A-company-bids-on-two-separate-contracts.lesson>A company bids on two separate contracts</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Advanced-probability-problems-related-to-combinations.lesson>Advanced probability problems related to combinations</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Using-cumulative-sums-and-relevant-standard-functuions-to-solve-probs-on-Binom-Distr.lesson#google_vignette>Using cumulative sums and relevant standard functions to solve problems on Binomial Distributions</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Advanced-binomial-distribution-problems.lesson>Miscellaneous binomial distribution problems</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Binomisal-distribution-problems-on-overbooking-flights.lesson>Binomial distribution problems on overbooking flights</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Advanced-probability-problems-on-binomial-distribution.lesson>Advanced probability problems on binomial distribution</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Accepting-rejecting-shipments-via-acceptance-procedures.lesson>Accepting/rejecting shipments via acceptance procedures</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Probabilistic-analysis-of-testing-procedures-in-health-care.lesson>Probabilistic analysis of testing procedures in health care</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Probabilistic-analysis-of-testing-procedures-in-industry.lesson>Probabilistic analysis of testing procedures in industry</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-problems-on-games.lesson>Probability problems on games </A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-problem-on-winning-many-rounds-game.lesson>Probability problem on winning a many-rounds game</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/A-Math-circle-level-probability-problem-on-a-lottery-game.lesson>A Math circle level probability problem on a lottery game</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Upper-league-entertainment-probability-problems.lesson>Upper league entertainment probability problems</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Upper-league-probability-problems-on-Stars-and-Bars-methodology.lesson>Upper league probability problems on Stars and Bars methodology</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Upper-league-problem-to-maximize-winning-in-a-game-with-20-sided-rolling-die.lesson>Upper league problem to maximize winning in a game with 20-sided rolling die</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Upper-league-problems-on-conditional-probability.lesson>Upper league problems on conditional probability</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Upper-league-geometric-probability-problems.lesson>Upper League geometric probability problems</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Problems-on-long-chains-of-related-events.lesson>Probability problems on long chains of related events</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Using-empirical-rules-to-determine-normal-distribution-probabilities.lesson>Using empirical rules to determine normal distribution probabilities</A> - <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Proba2.lesson>OVERVIEW of lessons on Probability, section 3</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/OVERVIEW-of-lessons-on-Probability.lesson>OVERVIEW of lessons on Probability</A> to navigate over all my lessons on Probability problems (section 1) in this site. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/OVERVIEW-of-my-Additional-lessons-on-Probability.lesson>OVERVIEW of my additional lessons on Probability</A> to navigate over all my lessons on additional Probability problems (section 2) in this site. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-II.
