Question

There are 90 counters in a bag. Each counter in the bag is either red or blue so that the number of red counters : the number of blue counters = 2 : 13. Li is going to put some more red counters in the bag so that the probability of taking at random a red counter from the bag is 1/3. Work out the number of red counters that Li is going to put in the bag?

Answer 1

To give a red counter a probability of 1/3 when drawn at random, Li needs to add 27 red counters to the current 12 in the bag, bringing the total number of counters to 117, and ensuring the desired probability.

Step-by-step explanation:

The original ratio of red to blue counters in the bag is 2:13, and there are 90 counters in total. To find the number of red and blue counters, you can write the ratio in terms of two unknowns, x and y, such that 2x + 13x = 90, as the total number equals red counters plus blue counters. Solving this equation yields x = 6, resulting in 12 red counters and 78 blue counters (2 × 6 = 12, and 13 × 6 = 78).

To achieve the desired probability of 1/3 for picking a red counter, we let the number of red counters that need to be added be 'n'. The new total count of counters will be 90+n, and the count of red counters will be 12+n. Set up the equation (12+n) / (90+n) = 1/3 and solve for n. This yields 36+n = 90+n, thereby finding n = -54.

However, since -54 is not a possible solution because we cannot have negative counters, we must have made an error. Re-evaluating the equation (12+n) / (90+n) = 1/3 and solving correctly, we get 36+3n = 90+n, which simplifies to 2n = 54, so n = 27. Therefore, Li must add 27 red counters to have a 1/3 probability of drawing a red counter.