Early in October COMSATS discovers that it can accommodate a few extra students. Enrolling those additional students would provide a substantial increase in revenue without increasing the operating costs of the university; that is, no new classes would have to be added. From experience the university knows that the frequency of enrollment given admission for all students is 40%.

A)      What is the probability that at most 4 students will enroll if the college offers admission to 10 more students?.

B)      If the frequency of enrollment given admission for all students was 70%, what is the probability that at least 12 out of 15 students will enroll?

 

A. The probability that at most 4 students will enroll if the college offers admission to 10 more students is 0.6331032576.

B. The probability that at least 12 out of 15 students will enroll is 0.296867927887048.

Step-by-step explanation

To solve this problems, note that the probability of an event (p) occurring exactly r times if there are n trials is given by the formula:

?(rn?)pr(1−p)n−r? .

A) We are looking for the probability that at most 4 students will enroll if the college offers admission to 10 more students (n=10). The phrase "probability that at most 4 students will enroll" means that:

• probability that no student will enroll (r=0),
• probability that exactly 1 student will enroll (r=1),
• probability that exactly 2 students will enroll (r=2),
• probability that exactly 3 students will enroll (r=3), and
• probability that exactly 4 students will enroll (r=4).

Base on the given, we know that p = 0.4; n = 10. We will then solve the probability for each r then take the sum of the probabilities to get the probability that at most 4 students will enroll. To do so, we will substitute the given values to the formula ?(rn?)pr(1−p)n−r? .

For r=0: ?(010?)0.40(1−0.4)10−0? = (1)(1)(0.0060466176)=0.0060466176

For r=1: ?(110?)0.41(1−0.4)10−1? = (10)(0.4)(0.010077696)=0.040310784

For r=2: ?(210?)0.42(1−0.4)10−2? =(45)(0.16)(0.01679616)=0.120932352

For r=3: ?(310?)0.43(1−0.4)10−3? =(120)(0.064)(0.0279936)=0.214990848

For r=4: ?(410?)0.44(1−0.4)10−4? =(210)( 0.0256)(0.046656)=0.250822656

Sum: 0.0060466176 + 0.040310784 + 0.120932352 + 0.214990848 + 0.250822656 = 0.6331032576

Hence, the probability that at most 4 students will enroll if the college offers admission to 10 more students is 0.6331032576.

B) We are looking for the probability that at least 12 out of 15 (n=15) students will enroll if the frequency of enrollment given admission for all students was 70% (p=0.7). The phrase "probability that at least 12 (out of 15) students will enroll" means that:

• probability that exactly 12 students will enroll (r=12),
• probability that exactly 13 students will enroll (r=13),
• probability that exactly 14 students will enroll (r=14), and
• probability that exactly 15 students will enroll (r=15).

We will solve the probability for each r then take the sum of the probabilities to get the probability that at least 15 students will enroll. To do so, we will substitute the given values to the formula ?(rn?)pr(1−p)n−r? .

For r=12: ?(1215?)0.712(1−0.7)15−12? = (455)(0.013841287201)(0.027)=0.170040213264285

For r=13: ?(1315?)0.713(1−0.7)15−13? = (105)(0.0096889010407)(0.09)=0.091560114834615

For r=14: ?(1415?)0.714(1−0.7)15−14? = (15)(0.00678223072849)(0.3)=0.030520038278205

For r=15:?(1515?)0.715(1−0.7)15−15? = (1)(0.004747561509943)(1)=0.004747561509943

Sum: 0.170040213264285 + 0.091560114834615 + 0.030520038278205 +0.004747561509943 = 0.296867927887048

Hence, the probability that at least 12 out of 15 students will enroll is 0.296867927887048.