A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:

Q: A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:

If last digit is either O or 5 then the number is divisible by 5. Case : 1 Last digit is 0. First three places can be selected by 9 × 9 × 9 = 729 ways. If c = 0 then three places can be selected by 9 × 8 × 1 = 72 If C ≠ 0 then 729 - 72 = 657 Fourth place has 8 choices = 657 × 8 = 5256 Total = 72 + 5256 = 5904 Case : 2 If C = 5 First place other than 5 then first three places can be filled in 8 × 8 × 1 = 64 If first place is 5 then first three places can be filled in 1× 9 × 1 = 9 ways. If third place is other than 5 then 729 - 64 - 9 = 656 ways. For fourth place has 8 choices. As per required condition = (64 + 9) × 9 + 656 × 8 = 5905 required probability $=\frac{5904+5905}{9\times10\times10\times10\times10}=\frac{11809}{90000}$ NOTE: Answer not matching with back answer.

M.C.Q (1 Marks)

GSEB - English Medium

A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:

A$\frac{1}{5}$
B$\frac{1}{5}\big(\frac{9}{10}\big)^3$
C$\big(\frac{3}{5}\big)^4$
D$\text{None of these}$

STD 12 Science
Maths
Probability
Exemplar

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