Difference between revisions of "2007 iTest Problems/Problem 21"
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== Problem == | == Problem == | ||
− | + | James writes down fifteen 1's in a row and randomly writes + or - between each pair of consecutive 1's. | |
+ | One such example is | ||
+ | <cmath>1+1+1-1-1+1-1+1-1+1-1-1-1+1+1.</cmath> | ||
+ | What is the probability that the value of the expression James wrote down is <math>7</math>? | ||
− | <math>\text{(A) } | + | <math>\text{(A) }0\qquad |
− | \text{(B) } 2\qquad | + | \text{(B) }\frac{6435 }{2^{14}}\qquad |
− | \text{(C) } | + | \text{(C) }\frac{6435 }{2^{13}}\qquad |
− | \ | + | \text{(D) }\frac{429}{2^{12}}\qquad |
− | + | \text{(E) }\frac{429}{2^{11}}\qquad | |
− | \text{( | + | \text{(F) }\frac{429}{2^{10}}\qquad |
− | + | \text{(G) }\frac{1}{15}\qquad | |
− | + | \text{(H) }\frac{1}{31}\qquad</math> | |
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− | \ | ||
− | |||
− | \text{( | ||
− | \ | ||
− | |||
− | \text{( | ||
− | |||
− | |||
− | \text{( | ||
− | \ | ||
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+ | <math>\text{(I) }\frac{1}{30}\qquad | ||
+ | \text{(J) }\frac{1}{29}\qquad | ||
+ | \text{(K) }\frac{1001 }{2^{15}}\qquad | ||
+ | \text{(L) }\frac{1001 }{2^{14}}\qquad | ||
+ | \text{(M) }\frac{1001 }{2^{13}}\qquad | ||
+ | \text{(N) }\frac{1}{2^{7}}\qquad | ||
+ | \text{(O) }\frac{1}{2^{14}}\qquad | ||
+ | \text{(P) }\frac{1}{2^{15}}\qquad</math> | ||
+ | |||
+ | <math>\text{(Q) }\frac{2007}{2^{14}}\qquad | ||
+ | \text{(R) }\frac{2007}{2^{15}}\qquad | ||
+ | \text{(S) }\frac{2007}{2^{2007}}\qquad | ||
+ | \text{(T) }\frac{1}{2007}\qquad | ||
+ | \text{(U) }\frac{-2007}{2^{14}}\qquad</math> | ||
== Solution == | == Solution == | ||
+ | |||
+ | In James’s expression, he is adding or subtracting <math>14</math> ones to one. Since the wanted result is <math>7</math>, he needs to add a total of <math>6</math> to <math>1</math>. This is achieved when he writes <math>10</math> plus signs and <math>4</math> minus signs, and there are <math>\binom{14}{10} = 1001</math> possible ways to do that. Since the total number of ways to put plus signs and minus signs between consecutive ones is <math>2^{14}</math>, the probability of getting a <math>7</math> is <math>\boxed{\textbf{(L) } \frac{1001}{2^{14}}}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{iTest box|year=2007|num-b=20|num-a=22}} | ||
+ | |||
+ | [[Category:Intermediate Probability Problems]] | ||
+ | [[Category:Intermediate Combinatorics Problems]] |
Latest revision as of 02:34, 14 June 2018
Problem
James writes down fifteen 1's in a row and randomly writes + or - between each pair of consecutive 1's. One such example is What is the probability that the value of the expression James wrote down is ?
Solution
In James’s expression, he is adding or subtracting ones to one. Since the wanted result is , he needs to add a total of to . This is achieved when he writes plus signs and minus signs, and there are possible ways to do that. Since the total number of ways to put plus signs and minus signs between consecutive ones is , the probability of getting a is .
See Also
2007 iTest (Problems, Answer Key) | ||
Preceded by: Problem 20 |
Followed by: Problem 22 | |
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