# 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?

4

Let's first work this forwards, then we can try working out our question (which will in essence be working backwards).

Since we certainly don't negative counters, we start with 4 red counters:

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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.

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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