Question 1125% of all students find this BS examdifficult. You select four random students. What is the probability thatexactly two of them find this exam difficult?1 point0.020.060.210.07
Question
Question 1125% of all students find this BS examdifficult. You select four random students. What is the probability thatexactly two of them find this exam difficult?1 point0.020.060.210.07
Solution
To solve this problem, we can use the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the number of combinations of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
In this case, we have:
- n = 4 (the number of students)
- k = 2 (the number of students who find the exam difficult)
- p = 0.25 (the probability that a student finds the exam difficult)
So, we can plug these values into the formula:
P(X=2) = C(4, 2) * (0.25^2) * ((1-0.25)^(4-2))
Calculating the combinations, C(4, 2) = 6.
So,
P(X=2) = 6 * (0.25^2) * ((1-0.25)^(4-2)) = 6 * 0.0625 * 0.5625 = 0.2109375
So, the probability that exactly two of the four students find the exam difficult is approximately 0.21.
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