Question

70% of the students applying to a university are accepted. Assume the requirements for a binomial experiment are satisfied for 10 applicants. a. What is the probability that among the next 10 applicants 8 or more will be accepted. b. What is the probability that among the next 10 applicants 4 or more will be accepted

Answer 1

Answer:

a) 0.3826

b) 0.9894

Step-by-step explanation:

We are given the following information:

We treat students accepted at university as a success.

P(students accepted at university) = 70% = 0.70

Then the number of students accepted at university follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

a) P( 8 or more will be accepted)

$P(x \geq 8) = P(x = 8) + P(x = 9) + P(X=10)\\\\= \binom{10}{8}(0.7)^8(1-0.7)^2 + \binom{10}{9}(0.7)^9(1-0.7)^1+ \binom{10}{10}(0.7)^{10}(1-0.7)^0\\\\= 0.2334 + 0.1210 + 0.0282\\= 0.3826$

b) P(4 or more will be accepted)

$P(x \geq 4) =1 - P(x = 0) - P(x = 1) - P(X=2)-P(x=3)\\\\=1 - ( \binom{10}{0}(0.7)^0(1-0.7)^{10} + \binom{10}{1}(0.7)^1(1-0.7)^9+ \binom{10}{2}(0.7)^{2}(1-0.7)^8+ \binom{10}{3}(0.7)^{3}(1-0.7)^7 )\\\\= 1 - 0.0106\\= 0.9894$