What is the probability that randomly selected rectangles will not contain both of the two red squares?

Question

What is the probability that randomly selected rectangles will not contain both of the two red squares? Note: Every square is also rectangle.

• All cases - (contain the upper red square or contain bottom red square)
• contain the upper red square or contain bottom red square
  = $\frac{54}{150} +\frac{48}{150} -\frac{24}{150} =\frac{78}{150}$.
• Then , $1 - \frac{78}{150} =\boxed{\frac{72}{150}}$

However, the given answer is $\boxed{\frac{24}{150}}$ . What am I missing ?

Answer 1

I think this boils down to ambiguities in the wording. Your math is correct, but maybe not for the interpretation that the problem setter intended.

1. Probability a rectangle contains neither red square, i.e that the number of red squares contained is zero. This is what you correctly calculated:

$$1-\frac{54}{150}-\frac{48}{150}+\frac{24}{150}=\frac{72}{50}$$

2. Probability of the complement of the event of containing both red squares. That is, the probability it contains either zero or one red square. This is one minus the probability that the rectangle contains both red squares, so $$ 1-\frac{24}{150}=\frac{126}{150} $$

I think that both of these situations are possible interpretations of the setup "What is the probability that randomly selected rectangles will not contain both of the two red squares?". It depends on whether you read this as "it is not true that the rectangle contains both red squares" or "for each red square, the rectangle does not contain it."

However, what is confusing is that $\frac{24}{150}$ fits with neither of these interpretations. Any way I slice it, it seems the book gave the wrong answer, and whether your answer is correct depends on how you interpret the phrasing.

Answer 2

In order for a randomly selected rectangle to include both red squares, two events must both happen.

$E_1$: the event that the lower left corner of the randomly selected rectangle is at or below the lower left corner of the lower red square, and at or to the left of the lower left corner of the lower red square.

Consequently, there are exactly $4$ choices for the lower left corner of the randomly selected rectangle.

$E_2$: the event that the upper right corner of the randomly selected rectangle is at or above the upper right corner of the upper red square, and at or to the right of the upper right corner of the upper red square.

Consequently, there are exactly $6$ choices for the upper right corner of the randomly selected rectangle.

This explains the enumeration of $6 \times 4.$

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I verified that there are in fact $150$ distinct rectangles that can be formed. Therefore, the probability that a randomly selected rectangle will contain both red squares is $\displaystyle ~\frac{24}{150}$.

Therefore, the probability that a randomly selected red square will not contain both red squares is

$$1 - \frac{24}{150} = \frac{126}{150}.$$

I suspect that there was a misprint in the problem's wording, either by the OP (i.e. original poster) or by the problem composer.

The alternative is that there is a mistake in my analysis. If there is one, I can't find it.