JEE Main & Advanced Mathematics Probability Question Bank Self Evaluation Test - Probability-I

  • question_answer
    A fair coin it tossed 2n times. The probability of getting as many heads in the first n tosses as in the last n is

    A) \[\frac{^{2n}{{C}_{n}}}{{{2}^{2n}}}\]

    B) \[\frac{^{2n}{{C}_{n-1}}}{{{2}^{n}}}\]

    C) \[\frac{n}{{{2}^{2n}}}\]

    D) None

    Correct Answer: A

    Solution :

    [a] The number of possible outcomes of 2n tosses is \[{{2}^{2n}}.\] There are \[^{n}{{C}_{r}}\]ways of getting r heads, with \[0\le r\le n,\] in n tosses. Therefore, the number of ways of getting r heads in both the first n and last n tosses is \[{{{{(}^{n}}{{C}_{r}})}^{2}}.\]Summing over all values of r. the number of favourable ways is \[{{{{(}^{n}}{{C}_{0}})}^{2}}+{{{{(}^{n}}{{C}_{1}})}^{2}}+{{{{(}^{n}}{{C}_{2}})}^{2}}+...+{{{{(}^{n}}{{C}_{n}})}^{2}}{{=}^{2n}}{{C}_{n}},\] So that the required probability is \[\frac{^{2n}{{C}_{n}}}{{{2}^{2n}}}.\]


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