Question

Mr. Johnson was trying to open a 12 digit lock set by his wife. He knew the first 8 digits. He also knew that the last 4 digits are 4, 5, 7, 8 but not the order in which they should be entered at the last 4 places. He keeps dialing a different number until the lock opens. What is the probability that the lock opens in the third attempt?

Answer 1

Answer with explanation:

Mr. Johnson knows the first 8 digit of a 12 digit lock.And ,He also knows, the last 4 digits are 4, 5, 7, 8, but not the exact arrangement of these four digits.

Arrangement of 4 digits, 4,5,7,8, if digits are not repeated = 1×2×3×4 or 4×3×2×1=24  ways

But,there is only one correct arrangement by which lock opens.

If we keep, first eight digit as equivalent to 1,and last four digit have 24 arrangement,then total number of arrangement of 12 digits =1×24=24

Probability of Success is equal to$\frac{1}{24}$

Probability of Failure ,P(F) is equal to $\frac{23}{24}$

Probability that the lock opens in the third attempt= F×F×S

   $=\frac{23}{24}\times\frac{23}{24} \times\frac{1}{24}\\\\=0.038266$

=0.038(Approx)