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STD 11 - 14. probability — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 14. probability4 Marks
MCQ
A three digit number is formed by using numbers $1, 2, 3$ and $4$. The probability that the number is divisible by $3$, is
  • A
    $\frac{2}{3}$
  • B
    $\frac{2}{7}$
  • ✓
    $\frac{1}{2}$
  • D
    $\frac{3}{4}$
✓

Answer

Correct option: C.
$\frac{1}{2}$
c
(c) Total number of ways to form the numbers of three digit with $1, 2, 3$ and $4$ are ${}^4{P_3} = 4\,! = 24$

If the numbers are divisible by three then their sum of digits must be $3, 6$ or $9$

But sum $3$ is impossible. Then for sum $6$, digits are $1, 2, 3$

Number of ways $ = 3\,!$

Similarly for sum $9$, digits are $2, 3, 4$. Number of ways =$3\, !$

Thus number of favourable ways $ = 3\,! + 3\,!$

Hence required probability $ = \frac{{3\,!\, + \,3\,!}}{{4\,!}} = \frac{{12}}{{24}} = \frac{1}{2}.$

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