Add the next number to this series?
60, 90, 108, 120, 128.5714*, 135, 140, 144, 147.2727*, _
(*5th and 9th numbers are rounded to four decimal places)
Answer: Well done to: David Mann, Polymer Business Development, France (thanks for the drawing below); John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Amparo Botella, Ismael Quesada SA, Spain; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, Larsen & Toubro Ltd, Kanchipuram, Tamil Nadu, India; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA.
As John Droogan noted, these are the angles inside a dodecagon: so, 10 x 180 / 12 = 150. We also liked Bharat B Sharma’s explanation (below) and similar approaches by Amparo Botella and Ramasubramanian P among others:
Sides Polygon Sum of all angles Interior angle l
3 Triangle 180° 60°
4 Quadrilateral 360° 90°
5 Pentagon 540° 108°
6 Hexagon 720° 120°
7 Heptagon 900° 128.5714…°
8 Octagon 1080° 135°
9 Nonagon 1260° 140°
10 Decagon 1440° 144°
11 Hendecagon 1620° 147.2727…°
12 Dodecagon 1800° 150
Question: A follow-on from last week’s ‘divisive question, we asked: There are 12 boys and 8 girls in the class. The teacher selects three children from the class at random to answer a question. What is the probability that the teacher chooses at least two boys?
Answer: A lot more clear-cut this week though only four correct answers suggest this type of question is proving a bit tricky for some. So extra well done to probably our best probability expert Michele Girardi, Scame Mastaf Spa, Suisio, Italy as well as to Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany, Bharat B Sharma, Sr V.P. product development & technical service (elastomers), Reliance Industries Ltd., India and P Ramasubramaniam , Larson & Toubro Ltd, India, who provided the following detailed analysis.
Probability of the teacher choosing at least two boys is the sum of probability of choosing all three boys and probability of choosing 2 boys and 1 girl.
Probability of choosing all 3 boys : P(BBB) = (12/20)*(11/19)*(10/18) = 0.193
Probability of choosing 2 boys and 1 girl: P(BBG,BGB,GBB) = (12/20)*(11/19)*(8/18) + (12/20)*(8/19)*(11/18)+(8/20)*(12/19)*(11/18) = (1056*3 )/6840 = 0.463
Probability of choosing at least 2 boys is : 0.193 + 0.463 = 0.656
P(at least 2 boys in a selection of 3) = (12C2 x 8C1 + 12C3 x 8C0)/20C3
= ((12×11/2) x (8/1) + (12x11x10/6)x1)/(20x19x18/6)
= (528 + 220)/1140 = 748/1140 = 0.656
In introducing herself. a new maths teacher tells her class that she has three children, before asking: if at least one of the children is a boy, what is the probability that she has three sons?
Answer: There were several different interpretations of this question, but the actual answer is 1/7 (or 14.3%), as all possible ways of the teacher having the three children must be considered – (ie BBB, BBG, BGB, GBB, GBG, GGB, BGG, but not GGG).
Extra well done to Michele Girardi, Scame Mastaf Spa, Suisio, Italy, who was the only one to get the correct answer, neatly explaining: There are 8 possible combinations of children, only one has no boys, and only one has 3 boys. So, the probability is 1/(8-1)
Tougher bonus question: Dividing time
If the digits 1 through to 9 are randomly arranged to make a number, what is the probability that that number is divisible by18?
Answer: 4/9 or 0.444. Well done to: Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Amparo Botella, Ismael Quesada SA, Spain; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, Larsen & Toubro Ltd, Kanchipuram, Tamil Nadu, India.
Again, Michele Girardi offered a detailed explanation: 18 = 2*3^2, so a number divisible by 18 must be divisible by 2 and 9. Numbers divisible by 2 must terminate by an even number, numbers are divisible by 9 if the sum of their digits is since: 1+2+3+4+5+6+7+8+9 = 45 (= 9×5), all the numbers created by this random arrangement are divisible by 9. Only the ones terminating by 2, 4, 6, 8, are also divisible by 2 and consequently by 18. The probability of this event is 4/9 = 14.3%. Then the total number of permutations is 9!,there are 8! permutations for each of the 4 even final digits, so the probability is 4*8! /9! = 4/
Complete the following set of factorials:
5! = 5 x 4 x 3 x 2 x 1=120, 4! = 4 × 3 × 2 × 1 = 24; 3! = ?; 2! = ?; 1! = ?; 0! = ?.
5! = 5 x 4 x 3 x 2 x 1=120; 4! = 4 × 3 × 2 × 1 = 24; 3! = 3 x 2 x 1 = 6; 2! = 2 x 1 = 2; 1! = 1 x 1 = 1; 0! = 1
Well done to all the following readers who avoided tripping over the final step: realising that 0! is not zero: Andrew Beasley, product analyst, Hankook Tyre UK Ltd, UK; David Mann, Polymer Business Development, France; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada: John Bowen, consultant, Bromsgrove, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Ramasubramanian P, L&T Rubber Processing Machinery, India; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria.
Among the many neat explanations provided were Andrew Beasley’s concise definition of n! as n! = n*(n-1)! for n > 0 and this longer one from Ramasubramanian P:
Mathematically, factorial of a positive integer n is the product of all the integers less than or equal to n and greater than or equal to 1. The significance of ‘factorial’ is that it represents the number of ways ‘n’ elements of a set can be distinctly arranged in a sequence. For example, for a set with three elements, say A,B,C the number of ways this set can be distinctly arranged is 6: ABC, ACB, BCA, BAC, CAB, CBA. Similarly, two elements can be arranged in two distinct ways. A single element can be arranged in only one distinct way. And a set with no elements can be arranged in one distinct way, which is again a NULL set. Hence 0! = 1 too.