We have a bag with 7 red discs and 5 blue discs. Two discs are drawn at random without replacement. We need to find the probability that at least one of the discs is red.

1. Problem Description

We have a bag with 7 red discs and 5 blue discs. Two discs are drawn at random without replacement. We need to find the probability that at least one of the discs is red.

2. Solution Steps

We can find the probability of the complementary event, which is the probability that none of the discs are red (i.e., both discs are blue). Then, we can subtract that probability from 1 to find the probability that at least one disc is red.

First draw:

The probability of drawing a blue disc on the first draw is the number of blue discs divided by the total number of discs:

$P(\text{first blue}) = \frac{5}{7+5} = \frac{5}{12}$

Second draw (given the first disc was blue):

Since we drew a blue disc on the first draw and did not replace it, there are now 4 blue discs and 7 red discs remaining, for a total of 11 discs.

The probability of drawing a blue disc on the second draw, given that the first disc was blue, is:

$P(\text{second blue} | \text{first blue}) = \frac{4}{11}$

The probability of drawing two blue discs is the product of these probabilities:

$P(\text{both blue}) = P(\text{first blue}) \times P(\text{second blue} | \text{first blue}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$

The probability of at least one red disc is the complement of the probability of both discs being blue:

$P(\text{at least one red}) = 1 - P(\text{both blue}) = 1 - \frac{5}{33} = \frac{33}{33} - \frac{5}{33} = \frac{28}{33}$

3. Final Answer

The probability that at least one of the discs is red is $\frac{28}{33}$.