From a sample set of 10 colored balls--5 green, 3 red and 2 yellow-- 5 are chosen at random WITH replacement. What is the probability that the result will be: green, red, yellow, red, green? Write your answer as a four place decimal number. p(G,R,Y,R,G)
since the balls are replaced, the draws are independent events. Just multiply all the probabilities:
5/10 * 3/10 * ...
.0045
To find the probability of the sequence "green, red, yellow, red, green" occurring while choosing 5 balls with replacement from the sample set, we first need to calculate the probability of each individual event.
1. Probability of choosing a green ball:
The sample set contains 5 green balls out of the total 10 balls. Therefore, the probability of selecting a green ball in each draw is 5/10, which can be simplified as 1/2.
2. Probability of choosing a red ball:
Similarly, the sample set contains 3 red balls out of 10. Therefore, the probability of selecting a red ball in each draw is 3/10.
3. Probability of choosing a yellow ball:
The sample set contains 2 yellow balls out of 10. Therefore, the probability of selecting a yellow ball in each draw is 2/10, which can be simplified as 1/5.
Since we are choosing with replacement, the probability of each event remains the same for each draw.
Now, let's calculate the probability of the specific sequence "green, red, yellow, red, green" occurring:
Probability = (Probability of green) * (Probability of red) * (Probability of yellow) * (Probability of red) * (Probability of green)
Probability = (1/2) * (3/10) * (1/5) * (3/10) * (1/2)
Calculating this expression, we get:
Probability = 0.009
Therefore, the probability of obtaining the sequence "green, red, yellow, red, green" while choosing 5 balls with replacement from the given sample set is 0.009, which can be written as a four-place decimal number.