4(х +Зу) = 2(3х-y) Solve for x
1 answer:
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Answer: 0.353
Step-by-step explanation:
We know that 42% of all the female students participate in 2 sports this season. (58% is the percent that did not)
This means that if we select a random student, there is a 0.42 probability that the student did participate in 2 sports this season.
Other thing that i will use is that, if we have a group of N objects, and we want to create a group of K objects out of the N, the number of combinations is:
Now, out of 5, we want to find the proability that, at least 3 of them, did participate in at least 2 sports this season.
The probability that all 5 of them participated, is equal to the product of the individual probabilities. (so N = 5, and K = 5)
P = (0.42^5)*1 = 0.013
Now, if only 4 of them did participate (N = 5, k = 4), the probability is equal to:
P = (0.42^4)*(0.58)*5!/4! = (0.42^4)*(0.58)*5 = 0.09
The probability for only 3 is: (N = 5 and K = 3)
P = (0.42^3)*(0.58^2)*5!/(3!*2!) = (0.42^3)*(0.58^2)*(5*4/2) = 0.25
Then the total probability is:
P = 0.013 + 0.09 + 0.25 = 0.353