Answer :
Answer:
There are 80 red counters.
Step-by-step explanation:
There are two really great ways to solve this. (Just focus on the one that works for you)
JUST DEAL WITH COUNTERS:
200 × .15 = 30 There are 30 pink counters.
200×.25= 50
There are 50 green counters.
Take away the pink and green.
200 - 30 - 50 = 120
There are 120 red and white counters.
There's twice as many red as white.
2x + x = 120
3x = 120
x = 40
There are 40 white counters.
There are 80 red counters.
OR, WORK WITH PROBABILITIES.
Pk + Grn + R + Wht = 100%
Use decimals, so the 100% is 1.00
.15 + .25 + 2x + x = 1
.4 + 3x = 1
3x = .6
x = .2
The probability of drawing a white counter is .2
The probability of drawing a red counter is .4
200 × .4 = 80
There are 80 red counters.
Final answer:
Given the other colors and their probabilities, the number of red counters in the bag can be calculated as 80 by taking into account that their drawing probability is twice that of the white counters, and the total number of counters in the bag is 200.
Explanation:
The problem states that the total number of counters in the bag is 200. The probability of drawing a pink counter is 0.15 and a green one is 0.25. If we multiply these probabilities by the total, we can find out exactly how many of each of these counters are in the bag. So for pink counters, we would calculate: 200 * 0.15 = 30 and for green counters: 200 * 0.25 = 50. This tells us that there are 30 pink and 50 green counters in the bag.
The problem also states that the probability of drawing a red counter is twice that of drawing a white one. Since the total probability must be 1, we can write the following equation to represent the white and red counters: 2*W + W = 1 - (0.15 + 0.25). Simplified, this is 3*W = 0.6, and therefore W = 0.6 / 3 = 0.2, or 20% probability for a white counter, and consequently 40% (or 0.4) for a red one.
Then, to find out how many red counters there are, we simply multiply the red counter probability by the total number of counters: 0.4 * 200 = 80 red counters
Learn more about Probability here:
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