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?

Answer 1

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#
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Answer 2

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

The probability of picking a red counter on the first draw is x10 \frac{x}{10} , as there are x x red counters out of the total of 10 counters.

After picking a red counter on the first draw, there are now x1 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 x19 \frac{x - 1}{9} .

According to the given information, the probability of picking two red counters in succession is 215 \frac{2}{15} .

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

x10×x19=215\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×x10×x19=90×21590 \times \frac{x}{10} \times \frac{x - 1}{9} = 90 \times \frac{2}{15}

9x(x1)=129x(x - 1) = 12

Expanding and simplifying:

9x29x=129x^2 - 9x = 12

9x29x12=09x^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 x , representing the number of red counters in the bag.

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Answer 3

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

The probability of picking a red counter on the first draw is x10 \frac{x}{10} , since there are x x red counters out of 10 total counters.

After picking one red counter, there will be x1 x - 1 red counters left in the bag, and the total number of counters will be 101=9 10 - 1 = 9 .

So, the probability of picking a red counter on the second draw, given that the first counter was red, is x19 \frac{x - 1}{9} .

According to the problem, the probability of picking two red counters in succession is 215 \frac{2}{15} .

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

x10×x19=215\frac{x}{10} \times \frac{x - 1}{9} = \frac{2}{15}

x(x1)90=215\frac{x(x - 1)}{90} = \frac{2}{15}

Cross-multiplying:

15x(x1)=2×9015x(x - 1) = 2 \times 90

15x215x180=015x^2 - 15x - 180 = 0

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

15x215x180=015x^2 - 15x - 180 = 0

Divide both sides by 15:

x2x12=0x^2 - x - 12 = 0

Now, we can factor the quadratic equation:

(x4)(x+3)=0(x - 4)(x + 3) = 0

x4=0x - 4 = 0 or x+3=0 x + 3 = 0

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

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

Therefore, there are 4 red counters in the bag.

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Answer from HIX Tutor

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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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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