Q: A swindler showed an honest man a six sided die.  If the man rolled a ONE, he wins, and gets back twice the amount of his bet. If not, the swindler keeps the bet.   “But…my chances are only one out of six,” retorted the man.  “True,” grinned the swindler, “But I’ll give you three tries to get a one.”  The man considered if I have 3 tries, each try has a 1/6 chance of winning, so my  chances of winning are  3/6 or 1/2. Is the bet really fair? If not, what are the chances of the man winning?

A:  91/216
You cannot just add 1/6 + 1/6 +1/6 so 1/2 is incorrect. The probability of not getting a 1 is 5/6 (there are 6 sides and the other possible outcomes are 2, 3, 4, 5 or 6).  The probability of no 1s in 3 throws is 5/6 x 5/6 x 5/6 = 125/216 which is the probability of the swindler winning.  So the probability of the man winning is 1 – 125/216  =   216/216 – 125/216 = 91/216.

transitiontoccregentsalgebraII1215update

The more things change, the more they stay the same.

The new Common Core Algebra II Regents will be supplemented by the old Algebra 2/Trig Regents this June 2016.  This new announcement comes on the heels of last week’s news of 2 more administrations of the Integrated Algebra in Feb and June 2016 for seniors. Info on Integrated Algebra Feb 2016 and June 2016.

Q: Find positive integers A, B and C such that 3^A + 5^B + 7^C (3 to the A power + 5 to the B power + 7 to the C) is a perfect square.  What is the smallest perfect square possible?

A: 81 is the smallest perfect square possible

List the powers of 3 5 and 7

3 9 27

5 25 125

7 49 343

If we combine 9 + 25 + 49 = 81