Our country has had 43 presidents. Write each president's last name on a card and place all 43 cards in a box. Note that several names ( e.g., Cleveland) occur twice. Now, randomly select 2 cards from the box.
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1. G. Washington 2. J. Adams 3. T. Jefferson 4. J. Madison 5. J. Monroe 6. J.Q. Adams 7. A. Jackson 8. M. Van Buren 9. W. H. Harrison 10. J. Tyler 11. J. Polk 12. Z. Taylor 13. M. Fillmore 14. F. Pierce 15. J. Buchanan |
16. A. Lincoln 17. A. Johnson 18. U.S. Grant 19. R. Hayes 20. J. Garfield 21. C. Arthur 22. G. Cleveland 23. B. Harrison 24. G. Cleveland 25. W. McKinley 26. T. Roosevelt 27. W. Taft 28. W. Wilson 29. W. Harding 30. C. Coolidge |
31. H. Hoover 32. F.D. Roosevelt 33. H. Truman 34. D. Eisenhower 35. J. F. Kennedy 36. L.B. Johnson 37. R. Nixon 38. G. Ford 39. J. Carter 40. R. Reagan 41. G.H.W. Bush 42. B. Clinton 43. G.W. Bush |
Solution by Drew Reyelts , Forest Hills Eastern High School
Of the 43 presidents, their last names begin with only 17 different letters.
Letters F , G , P , and W occur twice each. The probability of selecting one of these letters twice is (2/43)*(1/42)=2/1806. But there are 4 letters, so we multiply by 4. So, the probability of selecting any of these letters twice is 4*(2/1806)=8/1806.
Letters A , B , M , and R occur 3 times each. So, the probability of selecting any of these letters twice is 4*(3/43)*(2/42)=24/1806.
Letters J and T occur 4 times each. So, the probability of selecting any of these letters twice is 2*(4/43)*(3/42)=24/1806.
Letters C and H occur 5 times each. So, the probability of selecting any of these letters twice is 2*(5/43)*(4/42)=40/1806.
Add these probabilities together to get the total probability of selecting two names that begin with the same letter:
8/1806 + 24/1806 + 24/1806 + 40/1806 = 96/1806 = 5.3%.
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