If five digit numbers are formed from the digits $0,1,2,3$, 4 using every digit exactly only once, then the…

If five digit numbers are formed from the digits $0,1,2,3$, 4 using every digit exactly only once, then the probability that a randomly chosen number from those numbers is divisible by 4 , is
  1. $\frac{5}{16}$
  2. $\frac{3}{16}$
  3. $\frac{3}{8}$
  4. $\frac{7}{16}$

Solution

Total number of five digits number $=4 \cdot 4 \cdot 3 \cdot 2.1=96$ For number divisible by 4 contain Last two digit $04=3!=6$ Last two digit $40=3!=6$ Last two digit $20=3!=6$ Last two digit $12=2.2!=4$ Last two digit $32=2.2!=4$ Last two digit $24=2.2!=4$ $\therefore$ Total number divisible by 4 $=6+6+6+4+4+4=30$ Required probability $=\frac{30}{96}=\frac{5}{16}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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