Five digit numbers are formed using digits 1, 2, 3, 4, 5, 6 and 8 . probability that they have even

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14.Probability

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Five digit numbers are formed using the digits $1, 2, 3, 4, 5, 6$ and $8$. What is the probability that they have even digits at both the ends

A

$\frac{2}{7}$

B

$\frac{3}{7}$

C

$\frac{4}{7}$

D

None of these

Solution

(a) By using digits $1, 2, 3, 4, 5, 6$ and $8$, total $5$ digits numbers = $^7{P_5}$

And number of ways to form the numbers, they have even digit at both ends $ = 4 \times 3{ \times ^5}{P_3}$

$\therefore $ Probability $ = \frac{{4 \times 3{ \times ^5}{P_3}}}{{^7{P_5}}} = \frac{2}{7}$.

Standard 11

Mathematics

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