A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is

The digits available are $$3, 3, 4, 4, 4, 5, 5$$. The total number of distinct seven-digit arrangements is $$\frac{7!}{2! \cdot 3! \cdot 2!} = \frac{5040}{2 \cdot 6 \cdot 2} = 210$$.

For the number to be divisible by 2, the last digit must be even. The only even digit available is 4.

Fixing one 4 in the last position, the remaining six digits are $$3, 3, 4, 4, 5, 5$$. The number of distinct arrangements of these six digits is $$\frac{6!}{2! \cdot 2! \cdot 2!} = \frac{720}{8} = 90$$.

Therefore, the probability that the number is divisible by 2 is $$\frac{90}{210} = \frac{3}{7}$$.