A grid has 44 cells, of which 30 are blank. If one cell is chosen at random, what is the probability that it is blank?

Step-by-step solution

1. 1

   Identify the sample space and the favourable set. Every cell is equally likely, so the sample space has $44$ equally likely outcomes and the favourable set — the blank cells — has $30$ members. Equal likelihood is what licenses the counting formula in the next step; without it, counting would not be enough.

2. 2

   Apply the classical probability formula.

   $P(\text{blank}) = \frac{\text{number of blank cells}}{\text{total number of cells}} = \frac{30}{44}$

3. 3

   Reduce the fraction. Both numerator and denominator are even, and $\gcd(30,44) = 2$:

   $\frac{30}{44} = \frac{15}{22}$

   This is fully reduced because $15 = 3 \times 5$ and $22 = 2 \times 11$ share no prime factor.

4. 4

   Convert to a decimal and a percentage.

   $\frac{15}{22} = 0.681818\overline{18} \approx 68.18\%$

   The repeating block $18$ appears because $22 = 2 \times 11$ contains the prime $11$, which is not a factor of $10$.

5. 5

   Check with the complement. There are $44 - 30 = 14$ non-blank cells, so $P(\text{not blank}) = \tfrac{14}{44} = \tfrac{7}{22} \approx 0.3182$. The two probabilities sum to $\tfrac{15}{22} + \tfrac{7}{22} = 1$ exactly, as they must.