Question

A university found that 30% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course. Compute the probability that 2 or fewer will withdraw (to 4 decimals). Compute the probability that exactly 4 will withdraw (to 4 decimals). Compute the probability that more than 3 will withdraw (to 4 decimals). Compute the expected number of withdrawals.

Answer 1

Step-by-step explanation:

a) Compute the probability that 2 or fewer will withdraw

First we need to determine, given 2 students from the 20. Which is the probability of those 2 to withdraw and all others to complete the course. This is given by:

$(0.3)^2(0.7)^{18}$.

Then, we must multiply this quantity by

${20\choose2}=\frac{20!}{18!2!}=\frac{20\times19}{2}=190,$

which is the number of ways to choose 2 students from the total of 20. Therefore:

• the probability that exactly 2 students withdraw is $190(0.3)^2(0.7)^{18}$.

Following an analogous process we can determine that:

• The probability that exactly 1 student withdraw is ${20\choose1}(0.3)(0.7)^{19}=20(0.3)(0.7)^{19}.$

• The probability that exactly none students withdraw is ${20\choose 0}(0.7)^{20}=(0.7)^{20}.$

Finally, the probability that 2 or fewer students will withdraw is

$190(0.3)^2(0.7)^{18}+20(0.3)(0.7)^{19}+(0.7)^{20}=(0.7)^{18}(190(0.3)^2+20(0.3)(0.7)+(0.7)^2)\approx0.0355$

b) Compute the probability that exactly 4 will withdraw.

Following the process explained in a), the probability that 4 student withdraw is given by

${20\choose4}(0.3)^4(0.7)^{16}=\frac{20\times19\times18\times17}{4\times3\times2} (0.3)^4(0.7)^{16}=4845(0.3)^4(0.7)^{16}\approx 0.1304.$

c) Compute the probability that more than 3 will withdraw

First we will compute the probability that exactly 3 students withdraw, which is given by

${20\choose3}(0.3)^3(0.7)^{17}=\frac{20\times19\times18}{3\times2} (0.3)^3(0.7)^{17}=1140(0.3)^3(0.7)^{17}\approx 0.0716.$

Then, using a) we have that the probability that 3 or fewer students withdraw is 0.0355+0.0716=0.1071. Therefore the probability that more than 3 will withdraw is 1-0.1071=0.8929

d) Compute the expected number of withdrawals.

As stated in the problem, 30% of the students withdraw, then, the expected number of withdrawals is the 30% of 20 which is 6.