We assume each month is equally likely to be a student’s birthday month. i) The number of ways ten students can have birthdays in 10 differe

Question

We assume each month is equally likely to be a student’s birthday month. i) The number of ways ten students can have birthdays in 10 different months is: 12 · 11 · 10 . . . · 3 = 12!/2! ii) The number of ways 10 students can have birthday months is: 1210 iii) The probability that no two share a birthday month is: 12!/(2!.1210) = 0.00387 Is the solution showed correct?

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King 7 months 2021-08-03T00:54:15+00:00 1 Answers 6 views 0

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    2021-08-03T00:55:35+00:00

    Answer:

    The solutions are correct

    Step-by-step explanation:

    i) In a year there is a total of 12 months, if ten students can have birthdays in 10 different months, the number of ways this can happen is:

    Number of ways = 12 × 11 × 10 × . . . × 3 = \frac{12!}{2!}

    ii) The number of ways 1 student can have birthday months = 12¹, The number of ways 2 students can have birthday months = 12¹ × 12¹ = 12². Hence:

    The number of ways 10 students can have birthday months = 12¹⁰

    iii) The probability that no two share a birthday month = \frac{12!}{2!*12^{10}} = 0.00387

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