InterviewSolution
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InterviewSolution
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A coin is tossed `2n`times. The chance that the number of times one gets head is not equalto the number of times one gets tails is`((2n !))/((n !)^2)(1/2)^(2n)`b. `1-((2n !))/((n !)^2)`c. `1-((2n !))/((n !)^2)1/(4^n)^`d. none of theseA. `((2n!))/((n!)^(2))((1)/(2))^(2n)`B. `1-((2n!))/((n!)^(2))`C. `1-((2n!))/((n!)^(2))(1)/(4^(n))`D. None of these |
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Answer» Correct Answer - C The required probability is 1- probability of getting equal number of heads and tails `=1-""^(2n)C_(n)((1)/(2))^(n)((1)/(2))^(2n-n)` `=1-((2n)!)/((n!)^(2))xx(1)/(4^(n))` |
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