A bag contains 10 coloured counters. 2 are picked at random. Probability of them both being red is 2/15. Use algebra to work out how many of the 10 counters in the are red?

Question

Jonathan Allen · Accepted Answer

4

Let's first work this forwards, then we can try working out our question (which will in essence be working backwards). If I have 10 markers and 5 of them are red, on the first draw we'll have the probability of picking red is #5/10=1/2# And the second draw has a probability of picking red is #4/9#, giving the probability of picking two red: #1/2xx4/9=4/18=2/9# To work backwards then, we have a probability of #2/15#. To find the number of red counters, let's first set the initial number of red counters to be #r#. On the second draw, there is one less red counter, and so #r-1#. This gives: #r/10xx (r-1)/9=2/15# Now let's solve for #r#: #(r^2-r)/90=2/15=12/90# #:. r^2-r=12# #r^2-r-12=0# #(r-4)(r+3)=0# #=> r=4, -3# Since we certainly don't negative counters, we start with 4 red counters: #4/10xx3/9=2/5xx1/3=2/15 color(white)(000)color(green)root#

Elena Atkins · Answer

undefined Let $ x $ represent the number of red counters in the bag.

The probability of picking a red counter on the first draw is $ \frac{x}{10} $, as there are $ x $ red counters out of the total of 10 counters.

After picking a red counter on the first draw, there are now $ x - 1 $ red counters left in the bag, and 9 counters remaining in total.

So, the probability of picking another red counter on the second draw is $ \frac{x - 1}{9} $.

According to the given information, the probability of picking two red counters in succession is $ \frac{2}{15} $.

Using the multiplication rule for independent events, we multiply the probabilities of the two draws:

$$ \frac{x}{10} \times \frac{x - 1}{9} = \frac{2}{15} $$

To solve this equation, we can multiply both sides by 90 to eliminate the denominators:

$$ 90 \times \frac{x}{10} \times \frac{x - 1}{9} = 90 \times \frac{2}{15} $$

$$ 9x(x - 1) = 12 $$

Expanding and simplifying:

$$ 9x^2 - 9x = 12 $$

$$ 9x^2 - 9x - 12 = 0 $$

Now, we can solve this quadratic equation. Factoring or using the quadratic formula will give us the value(s) of $ x $, representing the number of red counters in the bag.

Nora Amin · Answer

undefined Let $ x $ represent the number of red counters in the bag.

The probability of picking a red counter on the first draw is $ \frac{x}{10} $, since there are $ x $ red counters out of 10 total counters.

After picking one red counter, there will be $ x - 1 $ red counters left in the bag, and the total number of counters will be $ 10 - 1 = 9 $.

So, the probability of picking a red counter on the second draw, given that the first counter was red, is $ \frac{x - 1}{9} $.

According to the problem, the probability of picking two red counters in succession is $ \frac{2}{15} $.

Using the multiplication rule for independent events, we multiply the probabilities of each event occurring:

$$ \frac{x}{10} \times \frac{x - 1}{9} = \frac{2}{15} $$

$$ \frac{x(x - 1)}{90} = \frac{2}{15} $$

Cross-multiplying:

$$ 15x(x - 1) = 2 \times 90 $$

$$ 15x^2 - 15x - 180 = 0 $$

Now, we can solve this quadratic equation for $ x $:

$$ 15x^2 - 15x - 180 = 0 $$

Divide both sides by 15:

$$ x^2 - x - 12 = 0 $$

Now, we can factor the quadratic equation:

$$ (x - 4)(x + 3) = 0 $$

$$ x - 4 = 0 $$ or $ x + 3 = 0 $

So, $ x = 4 $ or $ x = -3 $.

Since the number of red counters cannot be negative, the solution is $ x = 4 $.

Therefore, there are 4 red counters in the bag.