Question

There are 12 counters in a bag.

There is an equal number of red counters, blue counters and yellow
counters in the bag.

There are no other counters in the bag.

3 counters are taken at random from the bag.
(
a) Work out the probability of taking 3 red counters.

Answer 1

To work out the probability of taking 3 red counters from the bag, we need to calculate the ratio of the number of favorable outcomes (taking 3 red counters) to the total number of possible outcomes (taking any 3 counters).

Given that there are an equal number of red, blue, and yellow counters in the bag, let's assume there are n counters of each color.

The total number of possible outcomes is the number of ways to choose 3 counters out of the total 12 counters, which is given by the combination formula:

Total number of possible outcomes = C(12, 3) = 12! / (3!(12-3)!) = 12! / (3!9!) = (12 x 11 x 10) / (3 x 2 x 1) = 220

Now, to calculate the number of favorable outcomes (taking 3 red counters), we need to choose 3 counters out of the total n red counters:

Number of favorable outcomes = C(n, 3) = n! / (3!(n-3)!) = n! / (3! (n-3)!)

Since the number of red counters is equal to the number of blue and yellow counters, we have n red counters, n blue counters, and n yellow counters. Therefore, n = 12 / 3 = 4.

Plugging in the values, we get:

Number of favorable outcomes = C(4, 3) = 4! / (3! (4-3)!) = 4! / (3! x 1!) = (4 x 3 x 2) / (3 x 2 x 1) = 4

Therefore, the probability of taking 3 red counters is:

Probability = Number of favorable outcomes / Total number of possible outcomes = 4 / 220 = 1 / 55

So, the probability of taking 3 red counters is 1/55.

Answer 2

`(1)/(55)`

Explanation:

If there is a total of 12 counters in the bag, and there is an equal number of red counters, blue counters and yellow counters, then the bag contains:

• 4 red counters
• 4 blue counters
• 4 yellow counters

`\boxed{\sf Probability\:of\:an\:event\:occurring = (Number\:of\:ways\:it\:can\:occur)/(Total\:number\:of\:possible\:outcomes)}`

The probability of the first counter being red is:

`\sf P(Counter\;1:Red)=(4)/(12)`

Now we have 3 red counters and a total of 11 counters left in the bag. So the probability of the second counter being red is:

`\sf P(Counter\;2:Red)=(3)/(11)`

Now we have 2 red counters and a total of 10 counter left in the bag. So the probability of the third counter being red is:

`\sf P(Counter\;3:Red)=(2)/(10)`

To find the probability of the first and second and third counter being red, multiply the individual probabilities:

`\sf Probability=(4)/(12)* (3)/(11)*(2)/(10)=(24)/(1320)`

Reduce the fraction to its simplest form by dividing the numerator and denominator by the greatest common factor, 24:

`\sf (24)/(1320)=(24 / 24)/(1320/ 24)=(1)/(55)`

Therefore, the probability of taking 3 red counters is 1/55.