The digits 4 comma 5 comma 6 comma 7 comma and 8 are randomly arranged to form a​ three-digit number. ​ (Digits are not​ repeated.) Find the

Question

The digits 4 comma 5 comma 6 comma 7 comma and 8 are randomly arranged to form a​ three-digit number. ​ (Digits are not​ repeated.) Find the probability that the number is even and greater than 800.

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Rylee 1 week 2021-10-11T19:42:53+00:00 1 Answer 0

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    2021-10-11T19:44:18+00:00

    Answer:

    0.1 or 1/10

    Step-by-step explanation:

    The digits 4, 5, 6, 7 and 8 are randomly arranged to form a three digit number, where the digits are not repeated.

    This is question of permutation.

    Imagine this sum as; there are 3 boxes(blank spaces for digits) and 5 different fruits(digits) are to be put in these boxes, where a box can hold a maximum of only 1 fruit. The number of such permutations are: ⁵P₃

    By formula (a! is factorial a):

    ᵃPₙ = \frac{a!}{(a-n)!}

    ⁵P₃ = \frac{5!}{(5-3)!}

    ⁵P₃ = \frac{5!}{2!}

    ⁵P₃ = \frac{5*4*3*2*1}{2*1}

    ⁵P₃= 60

    This is the total count of possible numbers that can be formed.

    Now, for a number to be greater than 800 and even; first digit should necessarily be 8. Last digit can be 4 or 6. Using these conditions, there are 6 possibilities. 854, 864, 874, 846, 856, 876 are the numbers.

    The probability that number is even and greater than 800 is:

    P=\frac{ favorable outcomes}{equally likely possible outcomes}

    P=\frac{6}{60}

    P=\frac{1}{10}

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27:3+15-4x7+3-1=? ( )