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A coin is tossed 2n times. The probability that the number of times one get head is not equal to number of times one gats tail is ______

(a) 1 – (2/4n)

(b) 1 – [{(2n)!}/{(n!)2}] ∙ (1/4n)

(c) 1 – [{(2n)!}/{(n!)2}]

(d) [{(2n)!}/{(n!)2}] ∙ (1/4n)

1 Answer

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Best answer

The correct option (b) 1 - [{(2n)!}/{(n!)2}] ∙ (1/4n)

Explanation:

coin is tossed 2n times

Required probability = 1 – [probability that number of Head equal to number of tail = n]

= 1 – [(2nCn)(1/2)n(1/2)n

= 1 – [{(2n!)/[(n!)(n!)]} (1/4n)

= 1 – [{(2n)!}/{(n!)2}] ∙ (1/4n).

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