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Answer :
Answer:
[tex]\dfrac{1}{55}[/tex]
Step-by-step 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
[tex]\boxed{\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}}[/tex]
The probability of the first counter being red is:
[tex]\sf P(Counter\;1:Red)=\dfrac{4}{12}[/tex]
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:
[tex]\sf P(Counter\;2:Red)=\dfrac{3}{11}[/tex]
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:
[tex]\sf P(Counter\;3:Red)=\dfrac{2}{10}[/tex]
To find the probability of the first and second and third counter being red, multiply the individual probabilities:
[tex]\sf Probability=\dfrac{4}{12}\times \dfrac{3}{11}\times\dfrac{2}{10}=\dfrac{24}{1320}[/tex]
Reduce the fraction to its simplest form by dividing the numerator and denominator by the greatest common factor, 24:
[tex]\sf \dfrac{24}{1320}=\dfrac{24 \div 24}{1320\div 24}=\dfrac{1}{55}[/tex]
Therefore, the probability of taking 3 red counters is 1/55.
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Rewritten by : Barada
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.
