To answer this
problem, we use the binomial distribution formula for probability:
P (x) = [n!
/ (n-x)! x!] p^x q^(n-x)
Where,
n = the
total number of test questions = 10
<span>x = the
total number of test questions to pass = >6</span>
p =
probability of success = 0.5
q =
probability of failure = 0.5
Given the
formula, let us calculate for the probabilities that the student will get at
least 6 correct questions by guessing.
P (6) = [10!
/ (4)! 6!] (0.5)^6 0.5^(4) = 0.205078
P (7) = [10!
/ (3)! 7!] (0.5)^7 0.5^(3) = 0.117188
P (8) = [10!
/ (2)! 8!] (0.5)^8 0.5^(2) = 0.043945
P (9) = [10!
/ (1)! 9!] (0.5)^9 0.5^(1) = 0.009766
P (10) = [10!
/ (0)! 10!] (0.5)^10 0.5^(0) = 0.000977
Total
Probability = 0.376953 = 0.38 = 38%
<span>There is a
38% chance the student will pass.</span>
Answer:
1/2^4
Step-by-step explanation:
Answer:
We have x^3^2 which is x^3*2 = x*6 times x*4 which is x*10
Hope this helped and brainliest please
The possible different outcomes are: c. 6 outcomes.
you can only count "e" once because even though it is in the bag twice, drawing it a second time would be the same outcame.
hope this help
